28．（本题满分12分）如图1，已知?ABC的三顶点坐标分别为A(?1,?1)，B(3,?1)，C(0,?4)，二次函数y = ax2 + bx+c恰好经过A、B、C三点． (1)求二次函数的解析式；
(2)如图1，若点P是?ABC边AB上的一个动点，过点P作PQ∥AC，交BC于点Q，连接CP，当?CPQ的面积最大时，求点P的坐标； (3)如图2，点M是直线y?x上的一个动点，点N是二次函数图像上的一动点，若 ?CMN构成以CN为斜边的等腰直角三角形，直接写出所有满足条件的点N的横坐标．
(图1) (图2)
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2017年太仓市初中毕业暨升学考试模拟试卷
数学参考答案及评分标准
一、选择题（每小题3分，共30分） 题号 答案 1 B 2 A 3 C 4 D 5 B 6 C 7 A 8 D 9 C 14．3
10 C 二、选择题（每小题3分，共24分） 11．x?1 12．1.155×107
13．ab(a?b)(a?b)
15．1 16．23 17．32?6 18．7 三、解答题（共11大题，共76分） 19．（本题共4分） 解：原式= 2+9?1 ··········································································································· 3分
=10 ····················································································································· 4分
20．(本题共5分) 解：原式===xx ····································································································· 1分 ?x2?1x?1xx?1? ···························································································· 2分
(x?1)(x?1)x1 ················································································································ 3分 x?1当x=2?1时，原式==12 ·························································································· 4分
2． ·············································································································· 5分 221．(本题共6分) 解：由①式得：x＜3．···································································································· 2分
由②式得：x?1． ··································································································· 4分 3∴不等式组的解集为：1····································································· 5分 ?x?3． ·3??∴不等式组的整数解为：1,2． ··········································································· 6分 ??22．（本题满分6分） (1)8 ··································································································································· 1分 (2)144? ·························································································································· 3分 (3) 树状图或列表法略． ······························································································· 5分
5第一组至少有1名选手被选中的概率为． ·························································· 6分
623．（本题共6分）
(1)证明：∵CD=2，且?ADC与?ABD的面积比为1:3．∴BD=3DC=6 ························ 1分
∴在?ADC与?ABD中，
BCAC··························· 3分 ??2，∠BCA=∠ACD． ·
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∴?ADC∽?BAC． ··························································································· 4分
ADAB(2)解：∵?ADC∽?BAC，∴=，又∵AB?8，AC?4,CD?2．
DCAC∴．AD=4 ······································································································· 6分
24．（本题共9分）
·········································· 1分 解：(1)设A款文具盒单价为x元，则B公司为x+5元． ·
由题意得：
300300??5． ······················································································ 2分 xx?5········································································································· 3分 解之得：x=15． ·
···················································································· 4分 经检验：x=15是方程的根． ·
∴购进一个A款文具盒、一个B款文具盒分别需要15元和20元． (2)设购入A款文具盒为y个，则购入B款文具盒为60?y个． ········································································ 5分 由题意得：15y?20(60?y)?1000． ······································································································ 6分 解之得：y?40． ························ 7分 又∵售完60个文具盒可获得利润为S=5y?10(60?y)?600?5y ································································· 8分 ∴当y?40时，S可取得最大值为400． ·答：应购入40个A款文具盒和20个B款文具盒可使销售利润最大，最大利润
········································································································· 9分 为400元． ·25．（本题共8分） 解：(1) ∵点 A(?2，m+4)，点B(6，m)在反比例函数y?k的图像上． xk?m?4????2 ． ································································································· 1分 ∴?k?m??6?······················································································ 3分 ∴解得：m=?1，k=?6． · (2)设过A、B两点的一次函数解析式为y=ax+b． ??2k?b?3?k??2． ∵A(?2,3)，B(6,?1)，∴?．解得：??6k?b??1?b?2??11······································ 5分 ∴过A、B两点的一次函数解析式为y??x?2． ·21∵过点M(a，0)作x轴的垂线交AB于点P，∴点P的纵坐标为：?a?2．
2又∵过点M(a，0)作x轴的垂线交y??6?6于点Q，∴点Q的纵坐标为：． xa166∴PQ?|?a?2?| ，|QM|?|?|．
2aa1624················································ 7分 又∵PQ=4QM且a<0，∴?a?2???．
2aa∴a2?4a?60?0．∴a??6或a?10．
········································································ 8分 ∵a?0．∴实数a的值为?6． ·
26．（本题共10分） 解：(1) 连接CO．
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······························································· 1分 ∵D为BC的中点，且OB=OC，∴OD⊥BC． ·
∵OB=OC，∴∠OBC=∠OCB．
························································· 2分 又∵∠OBC=∠OFC，∴∠OCB=∠OFC． ·
∵OD⊥BC，∴∠DCF+∠OFC=90?．
···························· 3分 ∴∠DCF+∠OCB=90?．即OC⊥CF，∴CF为⊙O的切线． ·
F(2) ①设⊙O的半径为r． ∵OD⊥BC 且∠ABC=30?．
11∴OD=OB=r．
22CEDAOHB又∵DE=1，且OE=OD+DE．
1················································································ 4分 ∴r?1?r，解得：r=2． 2②作DH⊥AB于Ｈ，在RT△ODH中，∠DOH=60?，OD=1． ∴DH=13，OH=． 225在RT△DAH中，∵AH=AO+OH=，∴由勾股定理：AD=7． 2∴sin?BAD?DH321??······························································ 6分 ． ·AD2714 (3)设⊙O的半径为r． ∵O、D分别为AB、BC中点，∴AC=2OD． 又∵四边形ACFD是平行四边形，∴DF=AC=2OD． ∵∠OBC=∠OFC，∠CDF=∠ODB=90?，∴△ODB∽△CDF． ∴ODBDODBD??···································· 8分 ，∴，解得：BD?2OD． ·CDDFBD2OD36r,BD?r． 33∴在Rt△OBD中，OB=r，∴OD?∴OH?r,DH?132r． 34 ∴在RT△DAH中，∵AH=AO+OH=r，∴由勾股定理：AD=2r． 3 ∴sin?BAD? 27．（本题共10分） DH2r1??． ········································································· 10分 AD32r3·················································································································· 2分 解：(1)37 ·
(2) ∵四边形中ABCD，AB//CD,BC?AB,AD?CD?8,AB?12．
则BC?43，则S四边形ABCD=403． 1? 当0?t?4时．
如图，则BM=12?3t，CN=2t．
ADNCNDCMBAMB1∴S四边形BCNM?(12?t)?43?23(12?t)．
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·········· 3分 ∵MN将四边形的面积分为相等的两个部分，∴23(12?t)?203． ·
··················································································································· 4分 ∴t=2． ·
2? 当4?t?8时，
如图，则AM=24?3t，AN=16-2t
1333·········································· 5分 ∴S三角形AMN=??(24?3t)?(16?2t)?(8?t)2． ·
222∵MN将四边形的面积分为相等的两个部分，∴∴t=8?230230，又∵4?t?8，∴t=8?． 3333(8?t)2?203． 2综上所述：t?2或t=8?
230········································································ 6分 ． ·3DNPBCQDNPAMBC (3) 1?当0?t?4时， 如图，则AM=3t，CN=2t． AAPAM31???． ∵AB//CD，则PCCN22M············································································ 7分 ∴不存在符合条件的t值． ·2? 当4?t?8时，如图，分别延长CD、MN交于点Q． 则AM=24?3t，AN=16?2t，DN=2t?8． ∵AB//CD，则QDDNDQ2t?8????DQ?3(t?4) ···························· 8分 AMAN24?3t16?2tAMAP24?3t152????t?． CQPC3t?429∴CQ?3t?4． ∵AB//CD，则综上可知：存在实数t?52使得AP:PC?1:2成立． ······································· 10分 928．（本题共12分） 解：(1)y?x2?2x?4． ································································································ 3分 t?3)，则AP=t+1，BP=3?t，三角形ABC的面积为6． (2)设点P(t,?1)(?1?∵PQ//AC，∴?BPQ??BAC． ∴S?BPQS?BAC?(BP23?t23?t23)?()， ∴S?BPQ?()?S?BAC?(t?3)2 5分 BA44813又∵S?PCB??BP?3?(3?t)．
2233933∴S?PCQ?S?PBC?S?PBQ??t2?t???(t?1)2?． 8分
84882∴t=1时，S?PCQ最大，此时点P(1,?1)． ····························································· 9分 (3) 所有满足条件的点N的横坐标为?4,1?5,1?5． ······································ 12分
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