2024年普陀区初三数学二模参考答案及评分说明
一、选择题：（本大题共6题，每题4分，满分24分）
1．(B)； 2．(C)； 3．(A)； 4．(C)； 5．(D)； 6．(B). 二、填空题：（本大题共12题，每题4分，满分48分） 7.
23xy； 32?3； y8. x?3； 9. 4.027?108 ； 12. y?x2等；
10. y?11．>；
13．6； 14．
2 ； 1115．315； 18．(?5,?1?16．2a?b；
2三、解答题
17．3.14；
11)． 2（本大题共7题，其中第19---22题每题10分，第23、24题每题12分，第25题14分，满分78分）
x+2x2x?2g?19．解：原式? ················································· （3分） x?x?2?2(x?2)?x?2?x1 ····································································· （2分） ?x?2x?2x?1. ··············································································· （1分） ?x?2?当x?2?2时，原式?2?2?1 ························································ （1分）
2?2?22?3 ···························································· （1分） 2 ??2?32. ························································· （2分） 220．解：由①得，x≥-2. ·········································································· （3分）
由②得，x＜3. ··········································································· （3分） ∴原不等式组的解集是?2≤x<3． ················································· （2分） 所以，原不等式组的整数解是?2、?1、0、1、2． ··························· （2分）
21．解：
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（1）∵DE⊥AB，∴?DEA?90?
又∵?DAB?45o，∴DE?AE． ······················································ （1分）
3DE3，∴······················ （1分） ?． ·4BE4设DE?3x，那么AE?3x，BE?4x．
在Rt△DEB中，?DEB?90?，tanB?∵AB?7，∴3x?4x?7，解得x?1． ··············································· （2分） ∴DE?3． ··················································································· （1分） （2） 在Rt△ADE中，由勾股定理，得AD?32． ····································· （1分）
同理得BD?5． ············································································· （1分） 在Rt△ABC中，由tanB?∴CD?
2834，可得cosB?．∴BC?． ················ （1分） 4553． ·················································································· （1分） 5CD2?． ····························································· （1分） AD102． 10∴cos?CDA?
即?CDA的余弦值为22．解：
（1）x?0的实数； ·················································································· （2分） （2）?1； ······························································································ （2分） （3）图(略)； ·························································································· （4分） （4）图像关于y轴对称； 图像在x轴的上方；
在对称轴的左侧函数值y随着x的增大而增大，在对称轴的右侧函数值y随着x的增大而减小；
函数图像无限接近于两坐标轴，但永远不会和坐标轴相交等． ···················· （2分） 23．证明：
（1）∵ AD∥BC，DE∥AB，∴四边形ABED是平行四边形． ···················· （2分）
∵FG∥AD，∴同理
FGCF． ······························································ （1分） ?ADCAEFCF ． ··········································································· （1分） ?ABCAFGEF得＝
ABAD
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∵FG?EF，∴AD?AB． ······························································· （1分） ∴四边形ABED是菱形． ··································································· （1分） （2）联结BD，与AE交于点H．
∵四边形ABED是菱形，∴EH?1·························· （2分） AE，BD⊥AE． ·
2得?DHE?90o ．同理?AFE?90o．
∴?DHE＝?AFE． ········································································ （1分） 又∵?AED是公共角，∴△DHE∽△AFE． ······································· （1分）
EHDE． ··············································································· （1分） ?EFAE1∴AE2?EFgED． ······································································· （1分） 2∴24．解：
1（1） 由直线y?kx?3经过点C?2,2?，可得k??. ····································· （1分）
217由抛物线y??x2?bx?的对称轴是直线x?2，可得b?1. ················· （1分）
421（2） ∵直线y??x?3与x轴、y轴分别相交于点A、B，
2∴点A的坐标是?6,0?，点B的坐标是?0,3?. ········································ （2分）
?9?∵抛物线的顶点是点D，∴点D的坐标是?2,?. ·································· （1分）
?2?∵点G是y轴上一点，∴设点G的坐标是?0,m?.
∵△BCG与△BCD相似，又由题意知，?GBC??BCD，
∴△BCG与△BCD相似有两种可能情况： ··········································· （1分） ①如果
3?m5BGBC，那么，解得m＝·· （1分） ＝＝1，∴点G的坐标是?0,1?. ·
5CBCD523?m5BGBC1?1?，那么，解得m＝，∴点G的坐标是?0,?. （1分） ＝＝52CDCB5?2?2②如果
?1?综上所述，符合要求的点G有两个，其坐标分别是?0,1?和?0,? ．
?2?9??9??（3）点E的坐标是??1,?或?2,?. ···················································· （2分+2分）
42????
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25．解：
（1）过点O作OH⊥CD，垂足为点H，联结OC．
在Rt△POH中，∵sinP＝，PO?6，∴OH?2．······························ （1分） ∵AB＝6，∴OC＝····································································· （1分） 3． · 由勾股定理得 CH?5． ·································································· （1分）
∵OH⊥DC，∴CD?2CH?25． ················································· （1分） （2）在Rt△POH中，∵sinP＝，PO ＝m，∴OH＝221313m． ·························· （1分） 3?m?在Rt△OCH中，CH＝9???． ···················································· （1分）
3??m??在Rt△O1CH中，CH2＝36??n??． ············································· （1分）
3??m?3n2?81??m?可得 36??n??＝9???，解得m＝． ······························ （2分）
332n????（3）△POO1成为等腰三角形可分以下几种情况：
● 当圆心O1、O在弦CD异侧时
2223n2?81①OP＝OO1，即m＝n，由n＝解得n＝9． ······························ （1分）
2n即圆心距等于⊙O、⊙O1的半径的和，就有⊙O、⊙O1外切不合题意舍去．（1分） ②O1P＝OO1，由(n?m2m2)?m2?()＝n， 333n2?81922，即，······························ （1分） ＝n＝15． ·m＝nn解得解得
2n33581?3n2● 当圆心O1、O在弦CD同侧时,同理可得 m＝．
2n81?3n29∵?POO1是钝角，∴只能是m?n，即n＝，解得n＝5． ········ （2分）
2n599综上所述，n的值为5或15．55
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（完整版）2024年普陀区初三数学二模试卷及参考答案评分标准
