德庆2012九年级数学第一次模拟题参考答案和评分标准1

2012年九年级数学第一次模拟题参考答案和评分标准一、D C B A C B A D C C二、11、5- 12、2 13、6 14、π2 15、210三、16．解：原式=121211++- ························································································ 4分 =2 ··············································································································· 6分17．解：⎩⎨⎧<-≥+②x ①　x 04203 解不等式①，得3-≥x ·········································································································· 2分 解不等式②，得2<x ············································································································ 4分 不等式①，②的解集如下图所示：（图略）··········································································· 5分 所以这个不等式组的解集为23<≤-x ··············································································· 6分18．解：（1）依题意得，43411=-······················································································ 2分 所以取出白球的概率是43······································································································· 3分 （2）6414318=⨯÷（个） ··································································································· 5分 所以袋中的红球有6个 ··········································································································· 6分19．解：（1）40 ······················································································································ 2分（2）8，51（每答对一个给1分） ························································································ 4分 （3）88 ···································································································································· 7分 20．解：原式＝ab a b ab a 22222+-+- ······································································ 2分＝2b ········································································································································ 4分 当3=b 时，原式＝()332= ·················································································· 7分21．证明：在矩形ABCD 中，AD ∥BC ，∠A=90° ····························································· 2分 因为AD ∥BC ，∠AEB=∠EBC ····························································································· 3分 因为BC =CE ，所以∠EBC=∠BEC························································································· 4分 所以∠AEB=∠BEC ··············································································································· 5分 又∠A=∠BFE=90°，BE=BE ······························································································ 6分 所以△ABE ≌△FBE . ············································································································ 7分22．（1）证明：∵四边形ABCD 是平行四边形，∴A Q ∥BC ················································ 1分 ∴∠QDP=∠C ， ····················································································································· 2分 又∠QPD=∠BPC ····················································································································· 3分∴△DQP ∽△CBP ················································································································· 4分（2）∵△DQP ≌△CBP ，∴DP=CP=21CD, ······································································· 6分 ∵AB=CD=8, ∴DP=4. ·········································································································· 8分23．解：（1）∵直线x y 21=与双曲线x k y =(k ＞0)交于A 、B 两点，且点A 的横坐标为4， ∴2421=⨯=y ··················································································································· 2分 ∴A 点的坐标为（4，2） ········································································································ 3分 ∴24k =， 8=k ················································································································ 5分 （2）∵8=k ，∴反比例函数的表达式为x y 8=································································ 6分 当2-=x 时，4-=y ············································································································ 7分 ∴点（－2,－4）在双曲线上 ·································································································· 8分24．（1）解：CM 与⊙B 相切，理由是 ················································································· 1分 ∵BC 是直径，∴90BMC ∠= ，·························································································· 3分 所以CM 是⊙B 的切线 ··········································································································· 4分（2）因为⊙A 的半径是2，⊙B 的半径为1，即BC=4，BM=1 ··············································· 5分 在Rt △BMC 中，根据勾股定理求得CM=15 ········································································· 7分 由题意可知△BMC ∽△MPC ，所以MC BM MP BC ⋅=⋅， ············································· 8分所以MP ＝415，根据垂径定理可得MN=215 ···································································· 10分 25．解：(1)根据题意设21(2)3y a x =-- ············································································ 1分 .∵抛物线过点(1,0), ∴2(12)30,a --=∴3a =. ································································ 2分 ∴213(2)3y x =-- ··············································································································· 3分(2)相同点:1)对称轴相同;2)与x 轴的2个交点坐标相同;3)都经过一,四象限 ······················ 5分不同点:1)开口方向不同;2)顶点坐标不同;3)图象所在的象限不同 ········································· 7分(3) ∵B(2,1), ∴AB 的中垂线为直线1y =- ······································································· 8分∴23(2)31x --=-, ∴2x =,∴C (2 ························································ 9分同理,D (2 ················································································································ 10分。