绝密★启用前 试卷类型：A
2015年临沂市初中学生学业考试试题
数 学
注意事项：
1．本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题），共8页，满分120分，考试时间120分钟．答卷前，考生务必用0.5毫米黑色签字笔将自己的姓名、准考证号、座号填写在试卷和答题卡规定的位置．考试结束后，将本试卷和答题卡一并交回．
2．答题注意事项见答题卡，答在本试卷上不得分．
第Ⅰ卷（选择题 共42分）
一、选择题（本大题共14小题，每小题3分，共42分）在每小题所给出的四个选项中，只有一项是符合题目要求的．
1
1．-的绝对值是
2(A)
1. 21(B) -.
2
(C) 2. (D) -2.
2．如图，直线a ∥b ，∠1 = 60°，∠2 = 40°，则∠3等于 (A) 40°. (B) 60°. (C) 80°. (D) 100°.
3．下列计算正确的是 (A) a 2+a 2=2a 4. (C) a 2⋅a 3=a 6.
(B) (-a 2b ) 3=-a 6b 3. (D) a 8÷a 2=a 4.
（第2题图）
b a
1
3
2
4．某市6月份某周内每天的最高气温数据如下（单位：℃）：24 26 29 26 29 32 29 则这组数据的众数和中位数分别是
(A) 29，29. (B) 26，26. (C) 26，29. (D) 29，32.
5．如图所示，该几何体的主视图是
（第5题图）
(A) (B)
(C) (D)
⎧-2x
6．不等式组⎨的解集，在数轴上表示正确的是
⎩
x -2≤0
－3 －2 －1 0 1 2
－3 －2 －1 0 1 2
(A)
(B)
－3 －2 －1 0 1 2
－3 －2 －1 0 1 2
(C)
(D)
7．一天晚上，小丽在清洗两只颜色分别为粉色和白色的有盖茶杯时，突然停电了，小丽只好把杯盖和茶杯随机地搭配在一起. 则其颜色搭配一致的概率是
(A)
1
. 4
(B)
1. 2
(C)
3
. 4
(D) 1.
8．如图A ，B ，C 是e O 上的三个点，若∠AOC =100o ，则∠ABC 等于 (A) 50°. (C) 100°.
(B) 80°. (D) 130°.
2
9．多项式mx 2-m 与多项式x 2-2x +1的公因式是 (A) x -1. (C) x 2-1.
(B) x +1. (D) (x -1).
B
（第8题图）
10．已知甲、乙两地相距20千米，汽车从甲地匀速行驶到乙地，则汽车行驶时间t （单位：小时）关于行驶速度v （单位：千米／小时）的函数关系式是
(A) t =20v .
(B) t =. (C) t =.
v 20
(D) t =.
v
11．观察下列关于x 的单项式，探究其规律：x ，3x 2，5x 3，7x 4，9x 5，11x 6，…. 按照上述规律，第2015个单项式是 (A) 2015x 2015.
(B) 4029x 2014. (C) 4029x 2015. (D) 4031x 2015.
12．如图，四边形ABCD 为平行四边形，延长AD 到E ，使DE =AD ，连接EB ，EC ，DB . 添加一个条件，不能使四边形DBCE ．．成为矩形的是
(A) AB =BE . (B) BE ⊥DC . (C) ∠ADB =90°. (D) CE ⊥DE .
A
C
B
（第13．要将抛物线y =x 2+2x +3平移后得到抛物线y =x 2，下列平移方法正确的是 12题图） (A) 向左平移1个单位，再向上平移2个单位. (B) 向左平移1个单位，再向下平移2个单位. (C) 向右平移1个单位，再向上平移2个单位. (D) 向右平移1个单位，再向下平移2个单位.
14．在平面直角坐标系中，直线y =－x ＋2与反比例函数y =与反比例函数y =
1
的图象有唯一公共点. 若直线y =-x +b x
1
的图象有2个公共点，则b 的取值范围是
x
(A) b ﹥2. (B) －2﹤b ﹤2.
(C) b ﹥2或b ﹤－2. (D) b ﹤－2.
（第14题图）
第Ⅱ卷（非选择题 共78分）
注意事项：
1．第Ⅱ卷分填空题和解答题．
2．第Ⅱ卷所有题目的答案，考生须用0.5毫米黑色签字笔答在答题卡规定的区域内，在试卷上答题不得分．
二、填空题（本大题共5小题，每小题3分，共15分） 15．比较大小：
“﹤”，“＝”，“﹥”）. 16．计算：-2=____________.
a +2a +2a
17．如图，在Y ABCD 中，连接BD ，AD ⊥BD , AB =4, sin A =
3
，则Y ABCD 的面积是________. 4
C
D C
A B
B
（第17题图） （第18题图）
18．如图，在△ABC 中，BD ，CE 分别是边AC ，AB 上的中线，BD 与CE 相交于点O ，则
OB
=_________. OD
19．定义：给定关于x 的函数y ，对于该函数图象上任意两点（x 1，y 1），（x 2，y 2）， 当x 1﹤x 2时，都有y 1﹤y 2，称该函数为增函数. 根据以上定义，可以判断下面所给的函数中，是增函数的有______________（填上所有正确答案的序号）.
1
① y = 2x ； ② y =-x ＋1； ③ y = x 2 (x ＞0) ； ④ y =-.
x
三、解答题（本大题共7小题，共63分） 20．（本小题满分7分）
计算：1) .
“保护环境，人人有责”，为了了解某市的空气质量情况，某校环保兴趣小组，随机抽取了2014年内该市若干天的空气质量情况作为样本进行统计，绘制了如图所示的条形统计图和扇形统计图（部分信息未给出）.
请你根据图中提供的信息，解答下列问题： （1）补全条形统计图；
（2）估计该市这一年（365天）空气质量达到“优”和“良”的总天数； （3）计算随机选取这一年内的某一天，空气质量是“优”的概率.
某市若干天空气质量情况条形统计图
某市若干天空气质量情况扇形统计图
轻微污染 轻度污染
优 良
污染
污染
良
优
污染
污染
量类别
重度污染 中度污染
（第21题图）
22．（本小题满分7分）
小强从自己家的阳台上，看一栋楼顶部的仰角为30°，看这栋楼底部的俯角为60°，小强家与这栋楼的水平距离为42m ，这栋楼有多高？
（第22题图）
C
如图，点O 为Rt △ABC 斜边AB 上的一点，以OA 为半径的⊙O 与BC 切于点D ，与AC 交于点E ，连接AD .
（1）求证：AD 平分∠BAC ；
（2）若∠BAC = 60°，OA = 2，求阴影部分的面积（结果保留 ）.
24．（本小题满分9分）
新农村社区改造中，有一部分楼盘要对外销售. 某楼盘共23层，销售价格如下：第八层楼房售价为4000元／米2，从第八层起每上升一层，每平方米的售价提高50元；反之，楼层每下降一层，每平方米的售价降低30元，已知该楼盘每套楼房面积均为120米2.
若购买者一次性付清所有房款，开发商有两种优惠方案： 方案一：降价8%，另外每套楼房赠送a 元装修基金； 方案二：降价10%，没有其他赠送.
（1）请写出售价y （元／米2）与楼层x （1≤x ≤23，x 取整数）之间的函数关系式；
（2）老王要购买第十六层的一套楼房，若他一次性付清购房款，请帮他计算哪种优惠方案更加合算.
（第23题图） A
B
如图1，在正方形ABCD 的外侧，作两个等边三角形ADE 和DCF ，连接AF ，BE . （1）请判断：AF 与BE 的数量关系是 ，位置关系是 ；
（2）如图2，若将条件“两个等边三角形ADE 和DCF ”变为“两个等腰三角形ADE 和DCF ，且EA=ED=FD=FC”，第（1）问中的结论是否仍然成立? 请作出判断并给予证明；
（3）若三角形ADE 和DCF 为一般三角形，且AE=DF，ED=FC，第（1）问中的结论都能成立吗？请直接写出你的判断.
B
B
E
C
C
F
F 图1
图2
（第25 题图）
26. （本小题满分13分）
在平面直角坐标系中，O
为原点，直线y =－2x －1与y 轴交于点A ，与直线y =－x 交于点B , 点B 关于原点的对称点为点C .
（1）求过A ，B ，C 三点的抛物线的解析式； （2）P 为抛物线上一点，它关于原点的对称点为Q .
①当四边形PBQC 为菱形时，求点P 的坐标； ②若点P 的横坐标为t （－1＜t ＜1），当t 为何值时，四边形PBQC 面积最大，并说明理由.
（第26题图） B
E
C
A
D D
D
备用图
x
参考答案及评分标准
说明：解答题给出了部分解答方法，考生若有其它解法，应参照本评分标准给分. 一、选择题（每小题3分，共42分）
题号 1 2 3 4 5 6 7 8 9 10 11 12 13 答案
A
C
B
A
D
C
B
D
A
B
C
B
D
二、填空题（每小题3分，共15分）
15．>； 16．a ； 17．； 18．2； 19．①③.
三、解答题
20．解：方法一：1)
1) 1) ] ························································· 1分 =2-1) 2················································································ 3分 =3-(2-1) ·
·············································································· 5分 =3-2+1 ·················································································· 6分
=. ································································································· 7分
方法二：1)
=2121-111⨯1 ·
··········· 3分 =32+1 ·
······························································· 5分 =. ······················································································································ 7分 21．解：（1）图形补充正确. ························································································· 2分
某市若干天空气质量情况条形统计图
14 C
（2）方法一：由（1）知样本容量是60，
∴该市2014年（365天）空气质量达到“优”、“良”的总天数约为：
⨯365=292（天）.······························································································· 5分 60
方法二：由（1）知样本容量是60，
∴该市2014年（365天）空气质量达到“优”的天数约为：
⨯365=73（天）. ······································································································· 3分 该市2014年（365天）空气质量达到“良”的天数约为：
⨯365=219（天）. ····································································································· 4分 60
∴该市2014年（365天）空气质量达到“优”、“良”的总天数约为：
73+219=292（天）. ········································································································· 5分 （3）随机选取2014年内某一天，空气质量是“优”的概率为：
=. ····························································································································· 7分 22．解：如图，α = 30°，β = 60°，AD = 42.
∵tan α=
BD CD
，tan β=， AD AD
∴BD = AD ·tan α = 42×tan30°
····································· 3分
CD ＝AD tanβ＝42×tan60°
＝
···················································· 6分
C
∴BC ＝BD ＋CD ＝
＝
因此，这栋楼高为
··························································································· 7分
23．（1）证明：连接OD . ∵BC 是⊙O 的切线，D 为切点，
∴OD ⊥BC . ············································ 1分 又∵AC ⊥BC ，
∴OD ∥AC ， ·········································· 2分 ∴∠ADO =∠CAD. ································· 3分 又∵OD =OA ，
∴∠ADO =∠OAD ， ········································································································ 4分 ∴∠CAD =∠OAD ，即AD 平分∠BAC. ·········································································· 5分 （2）方法一：连接OE ，ED . ∵∠BAC =60°，OE =OA ， ∴△OAE 为等边三角形， ∴∠AOE =60°， ∴∠ADE =30°.
又∵∠OAD =∠BAC =30，
2∴∠ADE =∠OAD ，
∴ED ∥AO ， ············································· 6分 ∴S △AED ＝S △OED ，
∴阴影部分的面积 = S扇形ODE = =π. ························································· 9分
3603方法二：同方法一，得ED ∥AO ， ················································································· 6分 ∴四边形AODE 为平行四边形，
A
B
A
B
∴S V AED =S V OAD =⨯2 ·················································································· 7分
又S 扇形ODE －S △O ED
==π ······························································ 8分
3603∴阴影部分的面积 = (S扇形ODE －S △O ED ) + S△A ED
=ππ. ························ 9分
3324．解：（1）当1≤x ≤8时，y ＝4000－30（8－x ） ＝4000－240＋30 x
＝30 x＋3760；····························································· 2分
当8＜x ≤23时，y ＝4000＋50（x －8）
＝4000＋50 x－400 ＝50 x＋3600.
（1≤x ≤8，x 为整数），
⎧30x +3760
∴所求函数关系式为y =⎨ （ 8 ＜ 23 ， x 为整数 ） . ····························· 4分 x ≤
50x +3600⎩（2）当x ＝16时， 方案一每套楼房总费用：
w 1＝120（50×16＋3600）×92%－a ＝485760－a ； ················································ 5分 方案二每套楼房总费用：
w 2＝120（50×16＋3600）×90%＝475200. ······························································ 6分 ∴当w 1＜w 2时，即485760－a ＜475200时，a ＞10560； 当w 1＝w 2时，即485760－a ＝475200时，a ＝10560； 当w 1＞w 2时，即485760－a ＞475200时，a ＜10560. 因此，当每套赠送装修基金多于10560元时，选择方案一合算； 当每套赠送装修基金等于10560元时，两种方案一样；
当每套赠送装修基金少于10560元时，选择方案二合算. ············································· 9分 25．解：（1）AF =BE ，AF ⊥BE . ················································································· 2分 （2）结论成立. ················································································································ 3分 证明：∵四边形ABCD 是正方形，
B
∴BA =AD =DC ，∠BAD =∠ADC = 90°. 在△EAD 和△FDC 中， ⎧EA =FD , ⎪
⎨ED =FC , ⎪AD =DC , ⎩
∴△EAD ≌△FDC. ∴∠EAD =∠FDC.
∴∠EAD +∠DAB =∠FDC +∠CDA ，即∠BAE =∠ADF . ················································· 4分 在△BAE 和△ADF 中， ⎧BA =AD , ⎪
⎨∠BAE =∠ADF , ⎪AE =DF , ⎩
∴△BAE ≌△ADF.
∴BE = AF ，∠ABE =∠DAF. ··························································································· 6分 ∵∠DAF +∠BAF=90°， ∴∠ABE +∠BAF=90°，
∴AF ⊥BE . ······················································································································· 9分 （3）结论都能成立. ······································································································ 11分 ⎧y =-2x -1，⎧x =-1，26．解：（1）解方程组⎨得⎨
y =-x ，y =1. ⎩⎩
∴点B 的坐标为（-1，1）. ····························································································· 1分 ∵点C 和点B 关于原点对称，
∴点C 的坐标为（1，-1）. ····························································································· 2分 又∵点A 是直线y =-2x -1与y 轴的交点，
∴点A 的坐标为（0，-1）. ····························································································· 3分 设抛物线的解析式为y =ax 2+bx +c ，
⎧a -b +c =1，⎧a =1，⎪⎪∴⎨a +b +c =-1，解得⎨b =-1， ⎪c =-1. ⎪c =-1. ⎩⎩
∴抛物线的解析式为y =x 2-x -1. ························································································ 5分 （2）①如图1，∵点P 在抛物线上， ∴可设点P 的坐标为（m ，m 2-m -1）.
当四边形PBQC 是菱形时，O 为菱形的中心， ∴PQ ⊥BC ，即点P ，Q 在直线y = x 上，
∴m = m 2-m -1， ················································································································ 7分 解得m
··············································································································· 8分 ∴点P 的坐标为（
. ············································ 9分
x
图1 图2
②方法一：
如图2，设点P 的坐标为（t ，t 2 - t - 1）.
过点P 作PD ∥y 轴，交直线y = - x 于点D ，则D （t ，- t ）. 分别过点B ，C 作BE ⊥PD ，CF ⊥PD ，垂足分别为点E ，F .
∴PD = - t -( t 2 - t -1) = - t 2 + 1，BE + CF = 2， ······························································ 10分 ∴S △PBC ＝
11
PD ·BE +PD ·CF 22
＝＝
1
PD ·（BE + CF ） 2
1
（- t 2 + 1）×2 2
＝- t 2 + 1. ··········································································································· 12分
∴S Y PBQC ＝-2t 2+2.
∴当t ＝0时，S Y PBQC 有最大值2. ··············································································· 13分 方法二：
如图3，过点B 作y 轴的平行线，过点C 作x 轴的平行线，两直线交于点D ，连接PD . ∴S △PBC ＝S △BDC -S △PBD -S △PDC
＝
111
×2×2-×2（t +1）-×2（t 2-t -1+1） 222
＝-t 2+1. ·············································································································· 12分
∴S Y PBQC ＝-2t 2+2.
∴当t ＝0时，S Y PBQC 有最大值2. ··············································································· 13分
x
x
图3 图4
方法三：如图4，过点P 作PE ⊥BC ，垂足为E ，作PF ∥x 轴交BC 于点F . ∴PE =EF .
∵点P 的坐标为（t ，t 2-t -1），
∴点F 的坐标为（-t 2+t +1，t 2-t -1）. ∴PF =-t 2+t +1-t =-t 2+1. ∴PE
（-t 2+1）. ···································································································· 11分 11
BC ·PE ＝×
（-t 2+1）
22
∴S △PBC ＝
＝-t 2+1. ·············································································································· 12分
∴S Y PBQC ＝-2t 2+2.
∴当t ＝0时，S Y PBQC 有最大值2.
绝密★启用前 试卷类型：A
2015年临沂市初中学生学业考试试题
数 学
注意事项：
1．本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题），共8页，满分120分，考试时间120分钟．答卷前，考生务必用0.5毫米黑色签字笔将自己的姓名、准考证号、座号填写在试卷和答题卡规定的位置．考试结束后，将本试卷和答题卡一并交回．
2．答题注意事项见答题卡，答在本试卷上不得分．
第Ⅰ卷（选择题 共42分）
一、选择题（本大题共14小题，每小题3分，共42分）在每小题所给出的四个选项中，只有一项是符合题目要求的．
1
1．-的绝对值是
2(A)
1. 21(B) -.
2
(C) 2. (D) -2.
2．如图，直线a ∥b ，∠1 = 60°，∠2 = 40°，则∠3等于 (A) 40°. (B) 60°. (C) 80°. (D) 100°.
3．下列计算正确的是 (A) a 2+a 2=2a 4. (C) a 2⋅a 3=a 6.
(B) (-a 2b ) 3=-a 6b 3. (D) a 8÷a 2=a 4.
（第2题图）
b a
1
3
2
4．某市6月份某周内每天的最高气温数据如下（单位：℃）：24 26 29 26 29 32 29 则这组数据的众数和中位数分别是
(A) 29，29. (B) 26，26. (C) 26，29. (D) 29，32.
5．如图所示，该几何体的主视图是
（第5题图）
(A) (B)
(C) (D)
⎧-2x
6．不等式组⎨的解集，在数轴上表示正确的是
⎩
x -2≤0
－3 －2 －1 0 1 2
－3 －2 －1 0 1 2
(A)
(B)
－3 －2 －1 0 1 2
－3 －2 －1 0 1 2
(C)
(D)
7．一天晚上，小丽在清洗两只颜色分别为粉色和白色的有盖茶杯时，突然停电了，小丽只好把杯盖和茶杯随机地搭配在一起. 则其颜色搭配一致的概率是
(A)
1
. 4
(B)
1. 2
(C)
3
. 4
(D) 1.
8．如图A ，B ，C 是e O 上的三个点，若∠AOC =100o ，则∠ABC 等于 (A) 50°. (C) 100°.
(B) 80°. (D) 130°.
2
9．多项式mx 2-m 与多项式x 2-2x +1的公因式是 (A) x -1. (C) x 2-1.
(B) x +1. (D) (x -1).
B
（第8题图）
10．已知甲、乙两地相距20千米，汽车从甲地匀速行驶到乙地，则汽车行驶时间t （单位：小时）关于行驶速度v （单位：千米／小时）的函数关系式是
(A) t =20v .
(B) t =. (C) t =.
v 20
(D) t =.
v
11．观察下列关于x 的单项式，探究其规律：x ，3x 2，5x 3，7x 4，9x 5，11x 6，…. 按照上述规律，第2015个单项式是 (A) 2015x 2015.
(B) 4029x 2014. (C) 4029x 2015. (D) 4031x 2015.
12．如图，四边形ABCD 为平行四边形，延长AD 到E ，使DE =AD ，连接EB ，EC ，DB . 添加一个条件，不能使四边形DBCE ．．成为矩形的是
(A) AB =BE . (B) BE ⊥DC . (C) ∠ADB =90°. (D) CE ⊥DE .
A
C
B
（第13．要将抛物线y =x 2+2x +3平移后得到抛物线y =x 2，下列平移方法正确的是 12题图） (A) 向左平移1个单位，再向上平移2个单位. (B) 向左平移1个单位，再向下平移2个单位. (C) 向右平移1个单位，再向上平移2个单位. (D) 向右平移1个单位，再向下平移2个单位.
14．在平面直角坐标系中，直线y =－x ＋2与反比例函数y =与反比例函数y =
1
的图象有唯一公共点. 若直线y =-x +b x
1
的图象有2个公共点，则b 的取值范围是
x
(A) b ﹥2. (B) －2﹤b ﹤2.
(C) b ﹥2或b ﹤－2. (D) b ﹤－2.
（第14题图）
第Ⅱ卷（非选择题 共78分）
注意事项：
1．第Ⅱ卷分填空题和解答题．
2．第Ⅱ卷所有题目的答案，考生须用0.5毫米黑色签字笔答在答题卡规定的区域内，在试卷上答题不得分．
二、填空题（本大题共5小题，每小题3分，共15分） 15．比较大小：
“﹤”，“＝”，“﹥”）. 16．计算：-2=____________.
a +2a +2a
17．如图，在Y ABCD 中，连接BD ，AD ⊥BD , AB =4, sin A =
3
，则Y ABCD 的面积是________. 4
C
D C
A B
B
（第17题图） （第18题图）
18．如图，在△ABC 中，BD ，CE 分别是边AC ，AB 上的中线，BD 与CE 相交于点O ，则
OB
=_________. OD
19．定义：给定关于x 的函数y ，对于该函数图象上任意两点（x 1，y 1），（x 2，y 2）， 当x 1﹤x 2时，都有y 1﹤y 2，称该函数为增函数. 根据以上定义，可以判断下面所给的函数中，是增函数的有______________（填上所有正确答案的序号）.
1
① y = 2x ； ② y =-x ＋1； ③ y = x 2 (x ＞0) ； ④ y =-.
x
三、解答题（本大题共7小题，共63分） 20．（本小题满分7分）
计算：1) .
“保护环境，人人有责”，为了了解某市的空气质量情况，某校环保兴趣小组，随机抽取了2014年内该市若干天的空气质量情况作为样本进行统计，绘制了如图所示的条形统计图和扇形统计图（部分信息未给出）.
请你根据图中提供的信息，解答下列问题： （1）补全条形统计图；
（2）估计该市这一年（365天）空气质量达到“优”和“良”的总天数； （3）计算随机选取这一年内的某一天，空气质量是“优”的概率.
某市若干天空气质量情况条形统计图
某市若干天空气质量情况扇形统计图
轻微污染 轻度污染
优 良
污染
污染
良
优
污染
污染
量类别
重度污染 中度污染
（第21题图）
22．（本小题满分7分）
小强从自己家的阳台上，看一栋楼顶部的仰角为30°，看这栋楼底部的俯角为60°，小强家与这栋楼的水平距离为42m ，这栋楼有多高？
（第22题图）
C
如图，点O 为Rt △ABC 斜边AB 上的一点，以OA 为半径的⊙O 与BC 切于点D ，与AC 交于点E ，连接AD .
（1）求证：AD 平分∠BAC ；
（2）若∠BAC = 60°，OA = 2，求阴影部分的面积（结果保留 ）.
24．（本小题满分9分）
新农村社区改造中，有一部分楼盘要对外销售. 某楼盘共23层，销售价格如下：第八层楼房售价为4000元／米2，从第八层起每上升一层，每平方米的售价提高50元；反之，楼层每下降一层，每平方米的售价降低30元，已知该楼盘每套楼房面积均为120米2.
若购买者一次性付清所有房款，开发商有两种优惠方案： 方案一：降价8%，另外每套楼房赠送a 元装修基金； 方案二：降价10%，没有其他赠送.
（1）请写出售价y （元／米2）与楼层x （1≤x ≤23，x 取整数）之间的函数关系式；
（2）老王要购买第十六层的一套楼房，若他一次性付清购房款，请帮他计算哪种优惠方案更加合算.
（第23题图） A
B
如图1，在正方形ABCD 的外侧，作两个等边三角形ADE 和DCF ，连接AF ，BE . （1）请判断：AF 与BE 的数量关系是 ，位置关系是 ；
（2）如图2，若将条件“两个等边三角形ADE 和DCF ”变为“两个等腰三角形ADE 和DCF ，且EA=ED=FD=FC”，第（1）问中的结论是否仍然成立? 请作出判断并给予证明；
（3）若三角形ADE 和DCF 为一般三角形，且AE=DF，ED=FC，第（1）问中的结论都能成立吗？请直接写出你的判断.
B
B
E
C
C
F
F 图1
图2
（第25 题图）
26. （本小题满分13分）
在平面直角坐标系中，O
为原点，直线y =－2x －1与y 轴交于点A ，与直线y =－x 交于点B , 点B 关于原点的对称点为点C .
（1）求过A ，B ，C 三点的抛物线的解析式； （2）P 为抛物线上一点，它关于原点的对称点为Q .
①当四边形PBQC 为菱形时，求点P 的坐标； ②若点P 的横坐标为t （－1＜t ＜1），当t 为何值时，四边形PBQC 面积最大，并说明理由.
（第26题图） B
E
C
A
D D
D
备用图
x
参考答案及评分标准
说明：解答题给出了部分解答方法，考生若有其它解法，应参照本评分标准给分. 一、选择题（每小题3分，共42分）
题号 1 2 3 4 5 6 7 8 9 10 11 12 13 答案
A
C
B
A
D
C
B
D
A
B
C
B
D
二、填空题（每小题3分，共15分）
15．>； 16．a ； 17．； 18．2； 19．①③.
三、解答题
20．解：方法一：1)
1) 1) ] ························································· 1分 =2-1) 2················································································ 3分 =3-(2-1) ·
·············································································· 5分 =3-2+1 ·················································································· 6分
=. ································································································· 7分
方法二：1)
=2121-111⨯1 ·
··········· 3分 =32+1 ·
······························································· 5分 =. ······················································································································ 7分 21．解：（1）图形补充正确. ························································································· 2分
某市若干天空气质量情况条形统计图
14 C
（2）方法一：由（1）知样本容量是60，
∴该市2014年（365天）空气质量达到“优”、“良”的总天数约为：
⨯365=292（天）.······························································································· 5分 60
方法二：由（1）知样本容量是60，
∴该市2014年（365天）空气质量达到“优”的天数约为：
⨯365=73（天）. ······································································································· 3分 该市2014年（365天）空气质量达到“良”的天数约为：
⨯365=219（天）. ····································································································· 4分 60
∴该市2014年（365天）空气质量达到“优”、“良”的总天数约为：
73+219=292（天）. ········································································································· 5分 （3）随机选取2014年内某一天，空气质量是“优”的概率为：
=. ····························································································································· 7分 22．解：如图，α = 30°，β = 60°，AD = 42.
∵tan α=
BD CD
，tan β=， AD AD
∴BD = AD ·tan α = 42×tan30°
····································· 3分
CD ＝AD tanβ＝42×tan60°
＝
···················································· 6分
C
∴BC ＝BD ＋CD ＝
＝
因此，这栋楼高为
··························································································· 7分
23．（1）证明：连接OD . ∵BC 是⊙O 的切线，D 为切点，
∴OD ⊥BC . ············································ 1分 又∵AC ⊥BC ，
∴OD ∥AC ， ·········································· 2分 ∴∠ADO =∠CAD. ································· 3分 又∵OD =OA ，
∴∠ADO =∠OAD ， ········································································································ 4分 ∴∠CAD =∠OAD ，即AD 平分∠BAC. ·········································································· 5分 （2）方法一：连接OE ，ED . ∵∠BAC =60°，OE =OA ， ∴△OAE 为等边三角形， ∴∠AOE =60°， ∴∠ADE =30°.
又∵∠OAD =∠BAC =30，
2∴∠ADE =∠OAD ，
∴ED ∥AO ， ············································· 6分 ∴S △AED ＝S △OED ，
∴阴影部分的面积 = S扇形ODE = =π. ························································· 9分
3603方法二：同方法一，得ED ∥AO ， ················································································· 6分 ∴四边形AODE 为平行四边形，
A
B
A
B
∴S V AED =S V OAD =⨯2 ·················································································· 7分
又S 扇形ODE －S △O ED
==π ······························································ 8分
3603∴阴影部分的面积 = (S扇形ODE －S △O ED ) + S△A ED
=ππ. ························ 9分
3324．解：（1）当1≤x ≤8时，y ＝4000－30（8－x ） ＝4000－240＋30 x
＝30 x＋3760；····························································· 2分
当8＜x ≤23时，y ＝4000＋50（x －8）
＝4000＋50 x－400 ＝50 x＋3600.
（1≤x ≤8，x 为整数），
⎧30x +3760
∴所求函数关系式为y =⎨ （ 8 ＜ 23 ， x 为整数 ） . ····························· 4分 x ≤
50x +3600⎩（2）当x ＝16时， 方案一每套楼房总费用：
w 1＝120（50×16＋3600）×92%－a ＝485760－a ； ················································ 5分 方案二每套楼房总费用：
w 2＝120（50×16＋3600）×90%＝475200. ······························································ 6分 ∴当w 1＜w 2时，即485760－a ＜475200时，a ＞10560； 当w 1＝w 2时，即485760－a ＝475200时，a ＝10560； 当w 1＞w 2时，即485760－a ＞475200时，a ＜10560. 因此，当每套赠送装修基金多于10560元时，选择方案一合算； 当每套赠送装修基金等于10560元时，两种方案一样；
当每套赠送装修基金少于10560元时，选择方案二合算. ············································· 9分 25．解：（1）AF =BE ，AF ⊥BE . ················································································· 2分 （2）结论成立. ················································································································ 3分 证明：∵四边形ABCD 是正方形，
B
∴BA =AD =DC ，∠BAD =∠ADC = 90°. 在△EAD 和△FDC 中， ⎧EA =FD , ⎪
⎨ED =FC , ⎪AD =DC , ⎩
∴△EAD ≌△FDC. ∴∠EAD =∠FDC.
∴∠EAD +∠DAB =∠FDC +∠CDA ，即∠BAE =∠ADF . ················································· 4分 在△BAE 和△ADF 中， ⎧BA =AD , ⎪
⎨∠BAE =∠ADF , ⎪AE =DF , ⎩
∴△BAE ≌△ADF.
∴BE = AF ，∠ABE =∠DAF. ··························································································· 6分 ∵∠DAF +∠BAF=90°， ∴∠ABE +∠BAF=90°，
∴AF ⊥BE . ······················································································································· 9分 （3）结论都能成立. ······································································································ 11分 ⎧y =-2x -1，⎧x =-1，26．解：（1）解方程组⎨得⎨
y =-x ，y =1. ⎩⎩
∴点B 的坐标为（-1，1）. ····························································································· 1分 ∵点C 和点B 关于原点对称，
∴点C 的坐标为（1，-1）. ····························································································· 2分 又∵点A 是直线y =-2x -1与y 轴的交点，
∴点A 的坐标为（0，-1）. ····························································································· 3分 设抛物线的解析式为y =ax 2+bx +c ，
⎧a -b +c =1，⎧a =1，⎪⎪∴⎨a +b +c =-1，解得⎨b =-1， ⎪c =-1. ⎪c =-1. ⎩⎩
∴抛物线的解析式为y =x 2-x -1. ························································································ 5分 （2）①如图1，∵点P 在抛物线上， ∴可设点P 的坐标为（m ，m 2-m -1）.
当四边形PBQC 是菱形时，O 为菱形的中心， ∴PQ ⊥BC ，即点P ，Q 在直线y = x 上，
∴m = m 2-m -1， ················································································································ 7分 解得m
··············································································································· 8分 ∴点P 的坐标为（
. ············································ 9分
x
图1 图2
②方法一：
如图2，设点P 的坐标为（t ，t 2 - t - 1）.
过点P 作PD ∥y 轴，交直线y = - x 于点D ，则D （t ，- t ）. 分别过点B ，C 作BE ⊥PD ，CF ⊥PD ，垂足分别为点E ，F .
∴PD = - t -( t 2 - t -1) = - t 2 + 1，BE + CF = 2， ······························································ 10分 ∴S △PBC ＝
11
PD ·BE +PD ·CF 22
＝＝
1
PD ·（BE + CF ） 2
1
（- t 2 + 1）×2 2
＝- t 2 + 1. ··········································································································· 12分
∴S Y PBQC ＝-2t 2+2.
∴当t ＝0时，S Y PBQC 有最大值2. ··············································································· 13分 方法二：
如图3，过点B 作y 轴的平行线，过点C 作x 轴的平行线，两直线交于点D ，连接PD . ∴S △PBC ＝S △BDC -S △PBD -S △PDC
＝
111
×2×2-×2（t +1）-×2（t 2-t -1+1） 222
＝-t 2+1. ·············································································································· 12分
∴S Y PBQC ＝-2t 2+2.
∴当t ＝0时，S Y PBQC 有最大值2. ··············································································· 13分
x
x
图3 图4
方法三：如图4，过点P 作PE ⊥BC ，垂足为E ，作PF ∥x 轴交BC 于点F . ∴PE =EF .
∵点P 的坐标为（t ，t 2-t -1），
∴点F 的坐标为（-t 2+t +1，t 2-t -1）. ∴PF =-t 2+t +1-t =-t 2+1. ∴PE
（-t 2+1）. ···································································································· 11分 11
BC ·PE ＝×
（-t 2+1）
22
∴S △PBC ＝
＝-t 2+1. ·············································································································· 12分
∴S Y PBQC ＝-2t 2+2.
∴当t ＝0时，S Y PBQC 有最大值2.