2018年漳州市初中毕业暨高中阶段招生考试
数学参考答案及评分标准
一、选择题（共10题，每题3分，满分30分） 题号 答案 1 B 2 D 3 A 4 D 5 A 6 B 7 B 8 C 9 A 10 C 二、填空题（共6小题，每题4分，满分24分） 11．2 12．120 13．2018 14．增大 15．21 16．4
三、解答题（10大题，满分共96分） 17．解：原式=2?1?3 ··················································································· 6分 =0． ············································································································ 8分 18．解：情况一：
2121x?2x?1?x2?4x?1 ····················································· 2分 22=x?6x ······································································································ 5分 =x(x?6)． ·································································································· 8分 情况二：
2121x?2x?1?x2?2x ····································································· 2分 22=x?1 ········································································································· 5分 =(x?1)(x?1)． ···························································································· 8分 情况三：
2121x?4x?1?x2?2x ····································································· 2分 22=x?2x?1 ·································································································· 5分 =(x?1)2． ··································································································· 8分
A D
19．证明：四边形ABCD是等腰梯形， ?AB?DC，?B??C． ································· 4分 E为BC的中点， ?BE?EC． ·················································· 6分 B C E
（第19题） ?△ABE≌△DCE． ······································· 8分 20．（1）吉．（符合要求就给分） ······································································· 3分
（2）有多种画法，只要符合要求就给分． ··························································· 8分 C 21．（1）证明：连结OC， ································· 1分
AC?CD，?D?30°， 2 ??A??D?30° ············································ 2分
1 OA?OC， A D O B ??2??A?30°， ··········································3分
（第21题） ??1?60°，
??OCD?90°． ·························································································· 4分 ?CD是⊙O的切线． ····················································································· 5分 （2）?1?60°，
nπR60?π?3??π． ··································································· 7分 ?BC的长=180180答：BC的长为π． ························································································ 8分
y 22．（1）?1?x?3． ········································ 2分
y?x2?1
1 2（2）解：设y?x?1，则y是x的二次函数．
a?1?0，?抛物线开口向上． ··························· 3分
又
当y?0时，x?1?0，解得x1??1，x2?1． 4分
2?1 ?1 1 x ··· 6分 ?由此得抛物线y?x2?1的大致图象如图所示． ·
（第22题） 观察函数图象可知：当x??1或x?1时，y?0． ··············································· 7分
?x2?1?0的解集是：x??1或x?1． ···························································· 8分
23．（1）解法一：设甲种消毒液购买x瓶，则乙种消毒液购买(100?x)瓶． ·············· 1分 依题意，得6x?9(100?x)?780．
解得：x?40． ····························································································· 3分
． ····································································· 4分 ?100?x?100?40?60（瓶）
答：甲种消毒液购买40瓶，乙种消毒液购买60瓶． ············································· 5分 解法二：设甲种消毒液购买x瓶，乙种消毒液购买y瓶． ······································· 1分 依题意，得??x?y?100， ············································································· 3分
?6x?9y?780．解得：??x?40， ····························································································· 4分
y?60．?答：甲种消毒液购买40瓶，乙种消毒液购买60瓶． ············································· 5分 （2）设再次购买甲种消毒液y瓶，刚购买乙种消毒液2y瓶． ································· 6分 依题意，得6y?9?2y≤1200． ······································································ 8分 解得：y≤50． ···························································································· 9分 答：甲种消毒液最多再购买50瓶． ·································································· 10分
(，2)?26．（1）B（4，0），C0． ····································································· 2分
123x?x?2． ······················································································· 4分 22（2）△ABC是直角三角形． ··········································································· 5分
123证明：令y?0，则x?x?2?0．
22y??x1??1，x2?4．
?A(?1，0)． ································································································· 6分
解法一：?AB?5，AC?5，BC?25． ······················································ 7分
?AC2?BC2?5?20?25?AB2．
?△ABC是直角三角形． ················································································ 8分
COAO1，CO?2，BO?4，??? 解法二：AO?1BOOC2?AOC??COB?90°， ?△AOC∽△COB． ···················································································· 7分 ??ACO??CBO．
?CBO??BCO?90°，
??ACO??BCO?90°．即?ACB?90°． ?△ABC是直角三角形． ················································································ 8分
（3）能．①当矩形两个顶点在AB上时，如图1，CO交GF于H．
GF∥AB，
?△CGF∽△CAB． GFCH??． ················································ 9分 ABCO2解法一：设GF?x，则DE?x，CH?x，
52DG?OH?OC?CH?2?x．
5y D A O F G H C 图1
E B x 2?2??S矩形DEFG?x·?2?x???x2?2x
5?5?2?5?5=??x???． ····················································································· 10分
5?2?25时，S最大． 25?DE?，DG?1．
2△ADG∽△AOC，
当x?2
?ADDG11?，?AD?，?OD?，OE?2． AOOC22?1?0)． ··············································································· 11分 ?D??，0?，E(2，?2?10?5x． 210?5x555?S矩形DEFG?x·??x2?5x??(x?1)2?． ···································· 10分
2222?当x?1时，S最大．
5?DG?1，DE?．
2△ADG∽△AOC， ADDG11??，?AD?，?OD?，OE?2． AOOC22解法二：设DG?x，则DE?GF??1?0)． ··············································································· 11分 ?D??，0?，E(2，?2?y ②当矩形一个顶点在AB上时，F与C重合，如图2， DG∥BC，
?△AGD∽△ACB． GDAG??． BCAF解法一：设GD?x，?AC?5,BC?25，
D O A G C 图2
B G x ?GF?AC?AG?5?x． 2x?1?·?5????x2?5x ?S矩形DEFG?x2?2?1x?5=?2??25?．···················································································· 12分 2当x?5时，S最大．
?GD?5，AG?53522，?AD?AG?GD?．?OD?
222?3?································································································ 13分 ?D?，0? ·2??解法二：设
DE?，
AC?5，
BC?25，?GC?x，
AG?5?x．?GD?25?2x．
?S矩形DEFG?x·25?2x??2x2?25x
???5?5=?2?x?····················································································· 12分 ? ·???2?2?2?当x?5时，S最大， 253522．?AD?AG?GD?．?OD?.
222?GD?5，AG??3?································································································ 13分 0? ·?D?，2??综上所述：当矩形两个顶点在AB上时，坐标分别为??，（2，0）； 0?，
?1??2?当矩形一个顶点在AB上时，坐标为?，······················································ 14分 0? ·
?3??2?
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