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
2-的绝对值是
(A)
12. (B) 12
-.
(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) 2242a a a +=. (B) 2363()a b a b -=-. (C) 236a a a ?=.
(D) 824a a a ÷=.
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
b 1
3
2
(第2题图)
(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
(A) (B)
(C) (D)
6.不等式组2620
x x --???
<,
≤的解集,在数轴上表示正确的是
(A)
(B)
(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)
12
. (C)
3
4
. (D) 1.
8.如图A ,B ,C 是O e 上的三个点,若100AOC ∠=o ,则ABC ∠等于 (A) 50°. (B) 80°. (C) 100°.
(D) 130°.
9.多项式2mx m -与多项式221x x -+的公因式是 (A) 1x -. (B) 1x +. (C) 21x -.
(D) ()2
1x -.
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) 20t v =.
(B) 20t v =. (C) 20
v t =.
(D) 10t v
=.
-3 -2 -1 0 1 2
-3 -2 -1 0 1 2
-3 -2 -1 0 1 2
-3 -2 -1 0 1 2
O
A
B
C
(第8题图)
(第5题图)
11.观察下列关于x的单项式,探究其规律:x,3x2,5x3,7x4,9x5,11x6,….
按照上述规律,第2015个单项式是
(A) 2015x2015. (B) 4029x2014. (C) 4029x2015. (D) 4031x2015.
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.
13.要将抛物线223
y x x
=++平移后得到抛物线2
y x
=,下列平移方法正确的是
(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
1
y
x
=的图象有唯一公共点. 若直线y x b
=-+
与反比例函数
1
y
x
=的图象有2个公共点,则b的取值范围是
(A) b﹥2.
(B) -2﹤b﹤2.
(C) b﹥2或b﹤-2.
(D) b﹤-2. A
D
E
C
B
(第12题图)
(第14题图)
x
y
O
2
2
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:2_______3(填“﹤”,“=”,“﹥”). 16.计算:2422a a a a
-=++____________.
17.如图,在Y ABCD 中,连接BD ,AD BD ⊥, 4AB =, 3
sin 4
A =
,则Y ABCD 的面积是________.
(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 y = 2x ; ② y =-x +1; ③ y = x 2 (x >0); ④ 1
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:(321)(321)+--+.
O
B
C D E
A
B
C
D A
“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.
(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?
某市若干天空气质量情况扇形统计图
轻微污染 轻度污染
中度污染
重度污染 良
优
5%
某市若干天空气质量情况条形统计图 36
30
24 18 12 6 0
优 良
天数 空气质 量类别
重度 污染
轻微 污染
轻度 污染
中度 污染
12
36
3
2 1 C
A B
D α
β
(第22题图)
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. B
C
E
A
O
D (第23题图)
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.
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25题图)
B
A
E
F C
D 图1
备用图
B
A
C
D
图2 B
A
E
C
D
F (第26题图) O
x
y
A
C
B
21y x =--
y 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 14 答案
A
C
B
A
D
C
B
D
A
B
C
B
D
C
二、填空题(每小题3分,共15分)
15.>; 16.2a a -; 17.37; 18.2; 19.①③.
三、解答题
20.解:方法一:(321)(321)+--+
= [3(21)+-][3(21)--] ························································· 1分 =22(3)(21)--················································································ 3分 3(2221)=--+ ·
·············································································· 5分 32221=-+- ·················································································· 6分
22=. ································································································· 7分
方法二:(321)(321)+--+
22(3)323123(2)21131211=-?+?+?-+?-?+?-? ·
··········· 3分 363622321=-++-+-+- ·
······························································· 5分 22=. ······················································································································ 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
36 30 24 18
12 6 天数
12
36
3
2 6
(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:
123636529260
+?=(天).······························································································· 5分 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:
123657360
?=(天). ······································································································· 3分 该市2014年(365天)空气质量达到“良”的天数约为:
3636521960
?=(天). ····································································································· 4分 ∴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:
121.605
= ····························································································································· 7分 22.解:如图,α = 30°,β = 60°,AD = 42.
∵tan BD AD α=
,tan CD
AD
β=, ∴BD = AD ·tan α = 42×tan30°
= 42×
3
3
= 143. ····································· 3分 CD =AD tan β=42×tan60°
=423. ···················································· 6分
C
A B
D
α
β
∴BC =BD +CD =143+423
=563(m).
因此,这栋楼高为563m. ···························································································· 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∵1302OAD BAC ∠=∠=,
∴∠ADE =∠OAD ,
∴ED ∥AO , ············································· 6分 ∴S △AED =S △OED ,
∴阴影部分的面积 = S 扇形ODE = 60423603ππ??=. ························································· 9分
方法二:同方法一,得ED ∥AO , ················································································· 6分 ∴四边形AODE 为平行四边形,
B
C E
A
O
D
B
C
E
A
O
D
∴1S S 23 3.2AED OAD ==??=V V ·················································································· 7分
又S 扇形ODE -S △O ED =60423 3.3603ππ??-=- ······························································ 8分
∴阴影部分的面积 = (S 扇形ODE -S △O ED ) + S △A ED =223333ππ-+=. ························ 9分
24.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.
∴所求函数关系式为303760
503600x y x +?=?+? ····························· 4分 (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,
(1≤x ≤8,x 为整数),
(8<x ≤
23,x 为整数). B
A
∴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 26.解:(1)解方程组21y x y x =--??=-?,,得11.x y =-??=?
,
∴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 ,
∴111.a b c a b c c -+=??++=-??=-?,,解得111.a b c =??=-??=-?
,, ∴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 = 1±2. ··············································································································· 8分 ∴点P 的坐标为(1+2,1+2)或(1-2,1-2). ············································ 9分
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 =
12PD ·BE +1
2
PD ·CF O
x
y
P
A
C
B Q F
D E
21y x =--
y x
=- 21y x x =--
O
x
y
P
A
C
B
Q
21y x =--
y x
=- 21y x x =--
=1
2
PD ·(BE + CF ) =
1
2
(- t 2 + 1)×2 =- t 2 + 1. ··········································································································· 12分
∴S PBQC Y =-2t 2+2.
∴当t =0时,S PBQC Y 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
=
12×2×2-12×2(t +1)-1
2
×2(t 2-t -1+1) =-t 2+1. ·············································································································· 12分
∴S PBQC Y =-2t 2+2.
∴当t =0时,S PBQC Y 有最大值2. ··············································································· 13分
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),
O x
y
P
A
C
B
Q D
21y x =--
y x
=-
2
1y x x =--
O
x
y
P A
C
B
Q E 21y x =--
y x
=-
21y x x =--
F
∴点F的坐标为(-t2+t+1,t2-t-1). ∴PF=-t2+t+1-t=-t2+1.
∴PE=
2
2
(-t2+1). ····································································································11分
∴S△PBC=1
2
BC·PE=
1
2
×22×
2
2
(-t2+1)
=-t2+1. ··············································································································12分
∴S
PBQC
Y
=-2t2+2.
∴当t=0时,S
PBQC
Y
有最大值2.
https://www.wendangku.net/doc/907391495.html,
【精品】2020年山东省中考数学模拟试卷 含答案 一、选择题:本大题共10 小题,每小题 3 分,共30 分。在每小题给出的四个选项中,只有一项符合题目要求。 1.31-的值是() A.1 B.﹣1 C.3 D.﹣3 【解答】 解:31-=-1.故选B. 2.为贯彻落实觉中央、国务院关于推进城乡义务教育一体化发展的部署,教育部会同有关部门近五年来共新建、改扩建校舍186000000 平方米,其中数据186000000 用科学记数法表示是()A.1.86×107 B.186×106 C.1.86×108 D.0.186×109 【解答】解:将186000000 用科学记数法表示为:1.86×108.故选:C. 3.下列运算正确的是() A.a8÷a4=a2 B.(a2)2=a4 C.a2?a3=a6 D.a2+a2=2a4 【解答】解:A、a8÷a6=a4,故此选项错误; B、(a2)2=a4,故原题计算正确; C、a2?a3=a5,故此 选项错误;D、a2+a2=2a2,故此选项错误; 故选:B. 4.如图,点B,C,D 在⊙O 上,若∠BCD=130°,则∠BOD 的度数是 () A.50°B.60°C.80°D.100° 【解答】解:圆上取一点A,连接AB,AD,
∵点A、B,C,D 在⊙O 上,∠BCD=130°, ∴∠BAD=50°, ∴∠BOD=100°,故选:D. 5.多项式4a﹣a3 分解因式的结果是() A.a(4﹣a2)B.a(2﹣a)(2+a)C.a(a﹣2)(a+2)D.a(2﹣a)2 【解答】解:4a﹣a3 =a(4﹣a2)=a(2-a)(2+a).故选:B. 6..如图,在平面直角坐标系中,点A,C 在x 轴上,点C 的坐标为 (﹣1,0),AC=2.将Rt△ABC 先绕点 C 顺时针旋转90°,再向右平移 3 个单位长度,则变换后点 A 的对应点坐标是() A.(2,2)B.(1,2)C.(﹣1,2)D.(2,﹣1) 【解答】解:∵点 C 的坐标为(﹣1,0),AC=2, ∴点 A 的坐标为(﹣3,0), 如图所示,将Rt△ABC 先绕点 C 顺时针旋转90°,则点A′的坐 标为(﹣1,2), 再向右平移 3 个单位长度,则变换后点A′的对应点坐标为(2,2),故选:A. 7.在一次数学答题比赛中,五位同学答对题目的个数分别为7,5,3,5,10,则关于这组数据的说法不正确的是()
2019年山东省青岛市中考数学试卷 参考答案与试题解析 一、选择题(本大题共8小题,每小题3分,共24分)在每小题给出的四个选项中,只有一项是符合题目要求的. 1.(3分)﹣的相反数是() A.﹣B.﹣C.±D. 【分析】相反数的定义:只有符号不同的两个数互为相反数,0的相反数是0. 【解答】解:根据相反数、绝对值的性质可知:﹣的相反数是. 故选:D. 【点评】本题考查的是相反数的求法.要求掌握相反数定义,并能熟练运用到实际当中.2.(3分)下列四个图形中,既是轴对称图形,又是中心对称图形的是() A.B. C.D. 【分析】根据轴对称图形与中心对称图形的概念求解. 【解答】解:A、是轴对称图形,不是中心对称图形,故此选项错误; B、不是轴对称图形,是中心对称图形,故此选项错误; C、是轴对称图形,不是中心对称图形,故此选项错误; D、既是轴对称图形,又是中心对称图形,故此选项正确. 故选:D. 【点评】此题主要考查了中心对称图形与轴对称图形的概念.轴对称图形的关键是寻找对称轴,图形两部分折叠后可重合,中心对称图形是要寻找对称中心,旋转180度后两部分重合. 3.(3分)2019年1月3日,我国“嫦娥四号”月球探测器在月球背面软着陆,实现人类有史以来首次成功登陆月球背面.已知月球与地球之间的平均距离约为384000km,把384000km用科学记数法可以表示为() A.38.4×104km B.3.84×105km
C.0.384×10 6km D.3.84×106km 【分析】利用科学记数法的表示形式即可 【解答】解: 科学记数法表示:384 000=3.84×105km 故选:B. 【点评】本题主要考查科学记数法的表示,把一个数表示成a与10的n次幂相乘的形式(1≤a<10,n为整数),这种记数法叫做科学记数法. 4.(3分)计算(﹣2m)2?(﹣m?m2+3m3)的结果是() A.8m5B.﹣8m5C.8m6D.﹣4m4+12m5【分析】根据积的乘方以及合并同类项进行计算即可. 【解答】解:原式=4m2?2m3 =8m5, 故选:A. 【点评】本题考查了幂的乘方、积的乘方以及合并同类项的法则,掌握运算法则是解题的关键. 5.(3分)如图,线段AB经过⊙O的圆心,AC,BD分别与⊙O相切于点C,D.若AC=BD=4,∠A=45°,则的长度为() A.πB.2πC.2πD.4π 【分析】连接OC、OD,根据切线性质和∠A=45°,易证得△AOC和△BOD是等腰直角三角形,进而求得OC=OD=4,∠COD=90°,根据弧长公式求得即可. 【解答】解:连接OC、OD, ∵AC,BD分别与⊙O相切于点C,D. ∴OC⊥AC,OD⊥BD, ∵∠A=45°, ∴∠AOC=45°,
2020年湖南省中考数学模拟试题含答案 温馨提示: 1.本试卷包括试题卷和答题卡.考生作答时,选择题和非选择题均须作答在答题卡上,在本试题卷上作答无效.考生在答题卡上按答题卡中注意事项的要求答题. 2.考试结束后,将本试题卷和答题卡一并交回. 3.本试卷满分150分,考试时间120分钟.本试卷共三道大题,26个小题.如有缺页,考生须声明. 一、选择题(本大题共10个小题,每小题只有一个正确选项,请将正确选项填涂到答题卡 上.每小题4分,共40分) 1.如果a 与2017互为倒数,那么a 是( ) A . -2017 B . 2017 C . 20171- D . 2017 1 2.下列图形中,是中心对称图形的是( ) A. B. C. D. 3.下列计算正确的是( ) A . 6 33a a a =+ B . 33=-a a C . 5 23)(a a = D . 3 2a a a =?
4.人类的遗传物质是DNA,DNA是一个很长的链,最短的22号染色体与长达30000000个核苷酸,30000000用科学记数法表示为( ) A.3×107 B.30×104 C.0.3×107 D .0.3×10 8 5.如图,过反比例函数)0(>= x x k y 的图像上一点A 作 AB ⊥x 轴于点B ,连接AO ,若S △AOB =2,则k 的值为( ) A .2 B .3 C .4 D .5 6.下列命题:①若a<1,则(a﹣1) a a --=-111 ;②平行四边形既是中心对称图形又是轴对称图形;③9的算术平方根是3;④如果方程ax 2+2x+1=0有两个不相等的实数根,则实数a<1.其中正确的命题个数是( ) A.1个 B.2个 C.3个 D.4个 7.如图,AB ∥ CD,DE⊥ CE,∠ 1=34°,则 ∠ DCE的度数为( ) A.34° B.54° C.66° D.56° (第7题图) (第9题图) 8.一种饮料有两种包装,5大盒、4小盒共装148瓶,2大盒、5小盒共装100瓶,大盒与小盒每盒各装多少瓶?设大盒装x瓶,小盒装y瓶,则可列方程组( ) A. B. C. D . 9.如图,PA 、PB 是⊙O 的切线,切点分别为A 、B .若OA =2,∠P =60°,则?AB 的长为( )
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