最新-2018年中等学校招生贵港市统一考试数学参考答案

2018年中等学校招生贵港市统一考试
数学参考答案及评分标准
一、填空题
1．
13
2．20x -=（只要正确就给分） 3．110 4．增大 5．一 6．6 7．6 8．23 9．12 10．8 二、选择题：
11．C 12．B 13．A 14．C 15．B 16．A 17．D 18．D
三、解答题：
19．解：（1）原式21222
=⨯-⨯ ·································································· 3分 21=- ········································································································· 4分 1= ·············································································································· 5分
（2）22111
a a +-+ 21(1)(1)(1)(1)
a a a a a -=++-+- ·········································································· 3分 1(1)(1)a a a +=
+- ······························································································ 4分 11
a =- ········································································································ 5分 当3a =时，原式1111312
a ===--． ······························································· 6分 （不化简，直接代入不得分）
20．解：（1）
丹顶鹤
100 0.18 合计 2000
（填表每空1分，共3分，填图正确2分） ························································ 5分
（2）大熊猫． ······························································································ 7分
（3）答案如：①禁止乱捕滥杀野生动物．
②禁止人为破坏野生动物的生存环境．
（答案只要合理都得2分） ·························································· 9分
频数（学生人数） 1000 800 600 400
200 金丝猴 大熊猫 藏羚羊 丹顶鹤 动物名称
21．解：如图（只要画对其中两个，就给满分）．
22．解：（1）因为m y x =的图象经过点(12)A -，，(2)B n ，． 所以212
m m n ⎧=⎪⎪-⎨⎪=⎪⎩， ······························································································· 2分 解得：21m n =-=-，． ················································································ 4分
（2）由（1）得，点A B ，的坐标分别为(1
2)A -，，(21)B -，， 又因为一次函数y kx b =+，经过(12)A -，，(21)B -，， ········································ 5分
所以221k b k b -+=⎧⎨+=-⎩
， ··························································································· 7分 解得：11k b =-⎧⎨
=⎩，． ····························································································· 8分 所以一次函数的表达式为：1y x =-+． ····························································· 9分
23．解：（1）CBD A ∠=∠
（或CDB CBA ∠=∠或CD BC BC AC =，或CD BC BD BC AC AB
==等） ······························· 3分 （2）设CD x =，则2CA x =+． ····································································· 4分 若CBD CAB △∽△，且2AD =，3BC =，则
CD BC BC AC
= ··································································································· 5分 即323
x x =+ ······························································································· 6分 所以2230x x +-= ························································································ 7分 所以1213x x ==-，． ·
··················································································· 8分 经检验，1213x x ==-，都是原方程的解，但23x =-不符合题意，应舍去．
所以1CD x ==． ·························································································· 9分
24．解：设打折前A 商品每件x 元，B 商品每件y 元，依题意得： ··························· 1分 （2） （3）
（1）
63108584
x y x y +=⎧⎨+=⎩， ····························································································· 6分 解这个方程组得164
x y =⎧⎨=⎩， ················································································· 8分 所以打折前买5件A 商品和5件B 商品共用51654100⨯+⨯=（元）
所以100964-=（元） ·················································································· 9分 答：打折后买5件A 商品和5件B 商品比打折前少花了4元钱． ···························· 10分
25．解：（1）连接OD ． ················································································· 1分 AD 切O 于点D ，
OD AD ∴⊥．
90ADO ∴∠=， ·························································································· 2分 又30A ∠=，
60AOD ∴∠=，
1302
BED BCD AOD ∴∠=∠=∠=． ······························································ 3分 （2）DCE △是等边三角形．理由如下： ··························································· 4分 BC 为O 的直径且DE AC ⊥．
CE CD ∴=．
CE CD ∴=． ······························································································· 5分 BC 是O 的直径，
90BEC ∴∠=，
30BED ∠=，
60DEC ∴∠=，
DCE ∴△是等边三角形． ················································································ 6分
（3）O 的半径2R =． ∴直径4BC =
······························································································· 7分 由（2）知在Rt BEC △中，
sin 60CE BC =， ····························································································· 8分 sin 60CE BC ∴=
342
=⨯ ······································································································ 9分 23= ·
······································································································ 10分
26．解：（1）因为A C ，两点的横坐标分别为1，4，所以点(10)A ，． ·
······················ 1分 又点A B ，关于对称轴4x =对称，点(70)B ，． ·
··················································· 2分 （2）因为二次函数27y ax bx =+-的图象经过点(10)A ，，(70)B ，．
所以7049770a b a b +-=⎧⎨+-=⎩
， ···················································································· 4分 解得：18
a b =-⎧⎨=⎩， ····························································································· 6分 所以二次函数的表达式为287y x x =-+-． ······················································· 7分
（3）假设抛物线上存在点()P x y ，，使得45BAP ∠= ········································· 8分 ①当点P 在x 轴上方时有1x y -=，
2187x x x ∴-=-+-，
即2
760x x -+=．
解得：6x =或1x =（不合题意舍去）
所以，268675y =-+⨯-=． ∴点P 为(65)，
．···························································································· 9分 此时，130(71)51522
ABP S =⨯-⨯==△ ···························································· 10分 ②当点P 在x 轴的下方时，有1x y -=-
2187x x x ∴-=-+，
解得：8x =或1x =（不合题意舍去）
所以，2
88877y =-+⨯-=-． ∴点P 为(87)-，
． ······················································································ 11分 此时，142(71)72122
ABP S =⨯-⨯==△ ··························································· 12分。