普陀区2019学年第二学期初三质量调研数学试卷
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．(D)； 2．(C)； 3．(B)； 4．(D)； 5．(A)； 6．(C). 二、填空题：（本大题共12题，每题4分，满分48分） 7. 9a3； 10. 4； 13．m?1；
8. x??1；
9. x?0；
11．x?2y?0，x?y?0； 12. 80%b?a； 14．22； 17．y?10x；
15．25%； 18．4或
1r3r16．?a?b；
44三、解答题
8．（2分+2分） 5（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）
x1x2?2x?119．解：原式= ························································ （1分） ??x?1x2?1x?1x1(x?1)2??= ··················································· （2分） x?1(x?1)(x?1)x?1 =
x1 ········································································ （2分） ?x?1x?1x?1=. ················································································ （1分） x?13?1?1 ·························································· （1分）
3?1?13 ···························································· （1分） 3?2当x?3?1时，原式? ??23?3. ························································· （2分）
20．解：由①得，x≤2. ············································································ （3分）
由②得，x??1. ·········································································· （3分） ∴原不等式组的解集是?1 1 21．解：（1）∵一次函数y?2x?m的图像经过点A(?2,0)， ∴0?2?(?2)?m．解得 m?4． ························································· （2分） 同理可得 n??1． ············································································· （2分） 1∴一次函数的解析式分别是y?2x?4、y??x?1． 2∵当x?0时，y的值分别是4、?1， ∴B、C两点的坐标分别为(0,4)、(0,?1)． ··········································· （2分） （2）过点D作DH⊥y轴，垂足为点H． 11∵点D在直线y??x?1上，∴设点D的坐标为(a,?a?1)． 22 ∴DH?a． ∵B、C两点的坐标分别为(0,4)、(0,?1)，∴BC?5． ··························· （1分） 11由S△ABD?S△ABC?S△BCD，S△ABD?15，得?5?2??5?a?15． 22解得a?4． ······················································································ （2分） ∴点D的坐标为(4,?3)． ···························································································（1分） 22．解： 过点A分别作AH⊥MD、AG⊥CD，垂足分别为点H、G． ·······················（1分） 由题意得?ACD?45?，?BDC?90?，BD?2，CD?3.4， ∵AH⊥MD，∴?AHD?90?．同理，?AGD?90?． ∵?BDC?90?，∴四边形AGDH是矩形． ············································ （1分） ∴AH?GD，AG?HD． ·································································· （1分） 在Rt△ABH中，tan?ABH?由tan??AH， BH2AH2，得 ?． 5BH5设AH?2a，BH?5a， ····································································· （1分） ∴AG?5a?2，CG?3.4?2a． 在Rt△ACG中，?ACG?45?, AG5a?2····························································· （1分） ?1．即 ?1． · CG3.4?2a1解得 a? ····················································································· （1分） 5∴ 2 ∴AH?2，BH?1． ········································································ （2分） 5229． ·························· （2分） AH2?BH2?()2?12?55由勾股定理，得AB?∴显示屏的宽AB的长为23．证明： 29米． 5（1）∵四边形ABCD是平行四边形． ∴AO?CO． ············································································· （2分） ∵EA?EC，∴EO?AC． ·························································· （2分） ∴四边形ABCD是菱形． ······························································ （2分） （2）∵四边形ABCD是菱形, ∴?BAD?2?BAC，AD?CD． ························································· （2分） ∵?AEC?2?BAC，∴?BAD??AEC． ∵AB//CD，∴?BAD??CDF． ∴?AEC??CDF． ·········································································· （1分） 又∵?F??F， ∴△FCD∽△FAE． ··············································· （1分） ∴ CFCD． ·················································································· （1分） ?AFAE∴AE?CF?AF?CD． ∴EC?CF?AF?AD． ······································································· （1分） 24．解：（1）∵点A在x轴的正半轴上，与原点的距离为3， ∴点A的坐标是?3,0?. ········································································ （1分） 由抛物线y?ax2?4ax?3经过点A， 可得 9a?12a?3?0.解得 a?1. ························································ （1分） ∴抛物线的表达式是y?x2?4x?3. ··················································· （1分） 配方，得y?(x?2)2?1，∴顶点C的坐标是(2,?1). ····························· （1分） （2）∵点P是y轴下半轴上的一点，∴设点P的坐标为(0,y)，其中y??(BP?1)． 由点B(0,1)、点C(2,?1)，可得?CBP??CBD?45?，BC?22． ········· （1分） 3 设点D的坐标为(m,1)，由点D在抛物线y?x2?4x?3上，可得m2?4m?3?1， 解得m?2?2，m?2?2（舍）.∴BD?2?2. ·························· （1分） ∵△PBC与△CBD相似，且相似比不为1， ∴夹相等角的“两边对应成比例”只有一种情况： 即 BP22BPBC＝， 得 ， 解得BP?8?42， ···················· （1分） ＝BCBD222?2∴点P的坐标是(0,?7?42). ··························································· （1分） （3）∵直线y?1与y轴交于点B，∴点B的坐标是(0,1). ∴AB?10. 由A?3,0?、C(2,?1)，可得 AC?2. 得BC2?AC2?10，AB2?10.∴BC2?AC2?AB2. ∴?ACB?90?.∴tan?ABC?AC1················································ （1分） ?. · BC2过点E作EH?BD，垂足为点H． ∵?CBD??ABE，可得?ABC??EBH. ∴tan?EBH?1EH1. ∴?. 2BH2设EH?n，得BH?2n，点E的坐标为(2n,n?1)． 2得n?1?(2n)?4?(2n)?3. 解得n?2，n?1（舍）. ··································································· （2分） 4∴点E的坐标是?4,3?. ········································································ （1分） 25．解： （1）∵AD//BC，OF//BC，∴AD//OF//BC． ······································· （1分） ∵AO?OB，∴DF?FC． ······························································ （1分） ∴OF是梯形ABCD的中位线． 1∴OF?(AD?BC)． ······································································ （1分） 2∵AD?1，BC?5，∴OF?3． ························································ （1分） 即eO的半径长为3． 4 （2）设DC的中点为M，联结OM；过点D作DN?BC，垂足为点N． ∵DN?BC，∴?DNB??DNC?90?． 由AD//BC，?ABC?90?，得AD?BN?1，AB?DN?2r． ∴NC?4． 在Rt△DNC中，由勾股定理得 DC?2r2?4． ···································· （1分） ∵OM是梯形ABCD的中位线，∴OM//BC，OM?3． ························ （1分） ∵OM//BC，∴?OMH??C． ∵?OHM??DNC，∴△OHM∽△DNC． ········································· （2分） ∴ OHOM． ?DNDCy3． ?22r2r?43rr?42得 化简得y?． ·········································································· （2分） （3）∵OD?1?r2，DG?4?r2，∴DG?OD． ·································· （1分） △ODG成为等腰三角形只有下列两种情况： ①OD＝OG，即1?r2?3，解得r?22． ······································ （1分） ②OG＝DG，即3?4?r2，解得r?5，此时⊙O与直线DC只有一个公共点，不符合题意，舍去． ···························································································· （2分） 综上所述，r的值为22． 5
上海普陀区2020年初三数学二模答案
