2010年九年级第一次模拟检测
数学试题参考答案及评分说明
一、选择题（每小题2分，共24分）
1． B 2．B 3．B 4．A 5．C 6．A 7．C 8．B 9．D 10．C 11．C 12．C 二、填空题（每小题3分，共18分）
13．3 14．2＜x＜3 15．15 16．－1 17．90 18．三、解答题
19．解：去分母：x?1?2x?x?1 ··················································································· 2分
化简：x?x?0，解得：x1?0，x2??1． ························································ 5分 检验：x??1不是原方程的解． ············································································· 7分
所以原方程的解为x?0． ······················································································· 8分 20．解：（1）∵(252+104+24)÷1000=38%,
∴这1000名小学生患近视的百分比为38%.????????????2分
（2）∵(263+260+37)÷56%=1000（人）,
∴本次抽查的中学生有1000人. ????????????????4分 （3）∵8×
221 25260=2.08（万人）， 1000104=1.04（万人）， 1000∴该市中学生患“中度近视”的约有2.08万人. ??????????6分 ∵10×
∴该市小学生患“中度近视”的约有1.04万人. ??????????8分
21．证明：（1）证法一：连接OD．????????????????????1分
∵点D为BC的中点，点O为AB的中点． ∴OD为△ABC的中位线
∴OD∥AC????????????????????????????2分
∴∠DEC=∠ODE
∵DE⊥AC，∴∠DEC=90°，∴∠ODE =90°?????????????3分 ∴DE⊥OD
∴DE是⊙O的切线????????????????????????4分
（2）连接AD，?AB为直径，??BDA?90°．
?DE?AC，??CED?90°
在Rt△CED中，cos?C?CE CD九年级数学一模 第 11 页 共10页
∴cos30°?53， ∴CD?10 ··········································································· 5分 CD?点D为BC的中点，?BD?CD?10 ?AC?AB，??B??C?30° ··········································································· 6分
在Rt△ABD中．cos?B?DB10203，cos?30°?，AB?， ·················· 7分 ABAB3∴⊙O的半径为
103．······························································································ 8分 3????????????????????????2分
22．解：（1）两．两．
（2）画图略 ······················································································································ 4分 （3）设直线EF所表示的函数解析式为y?kx?b. 把E(10,0),F(11,45)分别代入y?kx?b，得
?10k?b?0?k?45， 解得?????????????????6分 ???11k?b?45?b??450.······················································· 7分 ?直线EF所表示的函数解析式为y?45x?450. ·把y?30代入y?45x?450,得45x?450?30.
2?x?10．????????????????????????????9分
3答：10点40分骑车人与客车第二次相遇. ······························································· 10分 23．解：（1）DE=2AM???????????????????????????1分 （2）DE=2AM???????????????????????????2分 （3）DE与AM之间的数量关系为：DE=2AM?????????????3分 证明：延长AM到E，使ME=AM???????????????????4分 又∵BM=MC，∠AMC=∠BME
∴△BME≌△CMA
∴∠EBM=∠ACM????????????????????????5分 ∴∠ABE=∠ABM+∠EBM=∠ABM+∠ACM=180°－∠BAC ∵△ABE和△ACD是等腰直角三角形
∴AB=AE,AC=AD=BE,∠BAE=∠CAD=90° ∴∠DAE=180°－∠BAC
九年级数学一模 第 12 页 共10页
∴∠ABE=∠DAE?????????????????????????6分 ∴△ABE≌△ADE????????????????????????7分 ∴AE=DE
即DE=2AM???????????????????????????8分
（4）DE=2AM??????????????????????????10分 24．解：（1）?AFD??DCA（或相等）．········································································ 2分 （2）?AFD??DCA（或成立），理由如下： ···························································· 3分
由△ABC≌△DEF，得
AB?DE，BC?EF（或BF?EC）BC??DEFBAC?，EDF??，?A． ??ABC??FBC??DEF??CBF，??ABF??DEC． ······································ 4分
?AB?DE，?在△ABF和△DEC中，??ABF??DEC，
?BF?EC，??△ABF≌△DEC，?BAF??EDC． ······································································ 5分
∴∠BAC－∠BAF=∠EDF－∠EDC ∴∠FAC=∠CDF．
??AOD??FAC??AFD??CDF??DCA， ??AFD??DCA． ······································································································ 6分 （3）如图，BO?AD． ·································································································· 7分 由△ABC≌△DEF，点B与点E重合，
A 得?BAC??BDF，BA?BD．
?点B在AD的垂直平分线上，
G 且?BAD??BDA． ········································ 8分 F O C ??OAD??BAD??BAC， B(E)
D ?ODA??BDA??BDF，
??OAD??ODA．
?OA?OD，点O在AD的垂直平分线上． ······························································ 9分
······················································· 10分 ?直线BO是AD的垂直平分线，BO?AD． ·
?20?2(x?1)?2x?18 ，25．解：（1）y???30
（2）设利润为w
(1?x?6,x为整数)(6?x?11,x为整数)?????4分
112?2y?z?2x?18?(x?8)?12?x?14??88w???y?z?30?1(x?8)2?12?1(x?8)2?18?88?(1?x?6,x为整数)??8分
(6?x?11,x为整数)
九年级数学一模 第 13 页 共10页
121x?14，当x=5时，w最大?17（元）???????????????9分 8811w?(x?8)2?18，当x=11时，w最大?19（元）????????????11分
881综上知：在第11周进货并售出后，所获利润最大且为每件19元???????12分
8w?26．解：（1）在Rt?ADC中,AC?4,CD?3,?AD?5，
∴PE∥BC ∴△APE∽△ACD??????????????????1分 ∴
EAAPEAx 即???????????????????????3分
ADAD5455x，DE?5?x????????????????????4分 44∴EA?（2）?BC?5,CD?3,?BD?2，
当点Q在BD上运动x秒后，DQ＝2－1.25x,则
1157y??DQ?CP?(4?x)(2?1.25x)?x2?x?4????????6分
2282527即y与x的函数解析式为：y?x?x?4，
82自变量的取值范围是：0＜x<1.6????????????????????7分
(3)分两种情况讨论：
①当?EQD?Rt?时，显然有EQ=PC=4－x ∵EQ⊥AC ∴△EQD∽△ACD
?EQDQ4?x1.25x?2?,即?,解得 x?2.5 ACDC43解得 x?2.5??????????????????????????9分
②当?QED?Rt?时，
∵∠CDA=∠EDQ，∠QED=∠C=Rt∠ ∴△EQD∽△ACD
?EQDQ5(4?x)1.25x?2?,即?, CDDA125解得 x?3.1??????????????????????????11分
综上所述，当x为2.5秒或3.1秒时，?EDQ为直角三角形。??????12分
九年级数学一模 第 14 页 共10页