2016年静安区九年级数学一模试卷

静安区2015学年第一学期期末学业质量调研 九年级数学试卷

参考答案及评分说明2016.1

一、选择题：

1．D； 2．D； 3．A； 4．C； 5．C； 6．C． 二、填空题：

7．?8a6； 8x??2．； 9．x?4； 10．1?m?3； 11．（3, -8）； 12．（2, 5）； 13．

1115； 14．2； 15．?a?b； 16．4或无解； 17．； 18．． 32132三、解答题：

(x?2)(x?3)(x?3)219．解：原式＝ ························································· （4分） ?(x?2)(x?2)x(x?2)＝

(x?2)(x?3)x(x?2)? ···························································· （1分） 2(x?2)(x?2)(x?3)＝

x． ·············································································· （2分） x?3?3时，原式＝

33?3?11?3??1?3． ························ （3分） 2当x1?3220．解：x?33x??0, ·································································· （1分） 2233x2?x?, ········································································ （1分）

223339x2?x?()2??, ························································ （2分）

24216333(x?)2?, ······································································ （2分）

4162x?333 ·················································································· （2分） ??443?333?33，x2? ······························································· （2分） 44九年级数学 第 5 页 共 8 页

x1?

21．解：（1）∵直线y?∴a?4， x与反比例函数的图像交于点A（3，a）

34． ·············································· （1分） ?3=4，∴点的坐标A（3，4）

3k

设反比例函数解析式为y?， ·························································· （1分）

x

k12∴4?,k?12，∴反比例函数解析式为y?． ································ （1分）

3x过点B作BH⊥x轴，垂足为H， 由tan??∴m?BH1··············· （1分） ?，设BH=m，则OB=3m，∴B（3m，m） ·

OB312，m??2（负值舍去）， ···················································· （1分） 3m∴点B的坐标为（6，2）． ································································ （1分）

（2）过点A作AE⊥x轴，垂足为E，

························································ （1分） S?OAB?S?OAE?S梯形AEHB?S?OBH ·

111·································· （1分） AE?OE?(AE?BH)?EH?OH?BH ·

222111=?3?4?(4?2)?3??6?2?9． ·········································· （2分） 222=

22．解：延长PQ交直线AB于点H，由题意得．

由题意，得PH⊥AB，AB=30，∠PAH=26 .6°，∠PBH=45°，∠QBH=33.7°， 在Rt△QBH中，cot?QBH?BH?1.50，设QH=x，BH=1.5x， ············· （2分） QH在Rt△PBH中，∵∠PBH=45°，∴PH= BH=1.5x， ······························· （2分） 在Rt△PAH中，cot?PAH?AH······················· （2分） ?2.00，AH=2PH=3x， ·

PH∵AH–BH=AB，∴3x?1.5x?30，x?20． ········································· （2分） ∴PQ=PH–QH=1.5x?x?0.5x?10． ·················································· （1分） 答：该电线杆PQ的高度为10米． ···························································· （1分）

九年级数学 第 6 页 共 8 页

23．证明：（1）∵AE2?EF?EC，∴

AEEC． ······································· （1分） ?EFAE又∵∠AEF=∠CEA，∴△AEF∽△CEA． ····································· （2分） ∴∠EAF=∠ECA， ··································································· （1分） ∵AD=AC，∴∠ADC=∠ACD， ·················································· （1分） ∵∠ACD =∠DCE+∠ECA=∠DCE+∠EAF． ·································· （1分）

（2）∵△AEF∽△CEA，∴∠AEC=∠ACB． ··········································· （1分）

∵DA=DB，∴∠EAF=∠B． ·························································· （1分） ∴△EAF∽△CBA． ···································································· （1分）

AFEF． ·········································································· （1分） ?BAACAFEF∵AC=AD，∴． ··························································· （1分） ?BAAD∴

∴AF?AD?AB?EF． ······························································· （1分）

24．解：（1）∵直线y?1x?1与x轴、y轴分别相交于点A、B， 2∴A（–2，0）、B（0，1）．∴OA=2，OB=1． ······································· （2分） ∵CD//x轴，∴∠OAB=∠CDA，∵∠CDA=∠OCA，∴∠OAB=∠OCA．····· （1分） ∴tan∠OAB=tan∠OCA， ·································································· （1分） ∴

OBOA12，∴?， ······························································ （1分） ?2OCOAOC∴OC?4，∴点C的坐标为（0，4）． ················································ （1分） （2）∵CD//x轴，∴

CDBC． ····························································· （1分） ?AOBOCD3∵BC=OC–OB=4–1=3,∴． ·············· （1分） ?，∴CD=6，∴点D（6，4）

21设二次函数的解析式为y?ax2?bx?4， ············································· （1分）

1?a??,??0?4a?2b?4,4 ………………（1分） ? ······························ （1分） ?34?36a?6b?4,??b?.2?∴这个二次函数的解析式是y??x2?

九年级数学 第 7 页 共 8 页

143···································· （1分） x?4． ·

2

25．解：（1）∵AD∥BC，∴∠DAC=∠ECB． ··················································· （1分）

又∵AD=CE，AC=CB，∴△DAC≌△ECB． ········································· （2分） ∴∠DCA＝∠EBC． ········································································ （1分） （2）过点E作EH⊥BC，垂足为H．AE=AC–CE=10?x．

CH?CE?cos?ACB?S?CBE43··········································· （1分） x，∴EH=x． ·

55113················································ （1分） ?BC?EH??10?x?3x．

225SAEFAE2?()， ······························· （1分） S?CEBCE∵AF//BC．∴△AEF∽△CEB，∴

3x2?60x?300y(10?x)2?∴，∴y?． ········································ （1分）

x3xx2定义域为0?x?55?5． ······························································· （1分） （3）由于∠DFC=∠EBC<∠ABC, 所以∠DFC不可能为直角．

（i）当∠DGF=90°时，∠EGC=90°，由∠GCE=∠GBC，可得△GCE∽△GBC．

∴tan?GBC?CGCEx···················································· （1分） ??．

GBCB103xEH3x5??在Rt△EHB中， tan?GBC?． ·················· （1分）

4BH50?4x10?x5x3x，解得x?0(舍去),或x?5． ?1050?4x3?52?60?5?300?15． ·∴S?AEF?············································· （1分）

5∴

（ii）当∠GDF=90°时，∠BCG=90°，

可得∠GEC=90°，∠CEB=90°， ················································· （1分）

33，S?AEF?． ······························ （1分） 223综上所述，如果△DFG是直角三角形，△AEF的面积为15或．

2可得BE=6，CE=8，AE=2，EF=

九年级数学 第 8 页 共 8 页