江苏省南京市建邺区2017年中考一模数学试题(含答案)

25．（本小题满分9分）

（1）解法一：∵关于x的一元二次方程x2－4mx＋4m2＋2m－4＝0有实数根，

∴△＝（－4m）2－4（4m2＋2m－4）＝－8m＋16≥0， ··································· 3分

∴m≤2． ······························································································· 4分 解法二：∵x2－4mx＋4m2＋2m－4＝0，∴（x－2m）2＝4－2m． ······················· 3分 ∴m≤2． ······························································································· 4分 （2）解法一：y＝x2－4mx＋4m2＋2m－4的顶点为M为（2m，2m－4）， ··········· 6分

∴MO2＝（2m）2＋（2m－4）2＝8（m－1）2＋8． ········································· 7分

∴MO长度的最小值为22． ····································································· 9分 解法二：y＝x2－4mx＋4m2＋2m－4的顶点为M为（2m，2m－4）， ·················· 6分

∴点M在直线l：y=x－4上， ···································································· 7分 ∴点O到l的距离即为MO长度的最小值22． ············································ 9分 26．（本小题满分12分）

解：（1）3000； ···························································································· 2分

（2）设汽车的速度为x km/h，则飞机的速度为8x km/h，根据题意得：

3000－24002400

－＝3， ············································································ 4分 x8x解之得：x＝100．

经检验，x＝100为原方程的解.则飞机的速度为8×100＝800 km/h．

答：飞机的速度为800 km/h． ···································································· 6分 （3）图略． ··························································································· 8分 当0≤x≤3，y1=800x．

当3

?3k＋b＝2400，?k＝100，

代入点（3，2400），（9，3000）得：?解得?

?9k＋b＝3000?b＝2100．

∴函数关系式为：y1＝100x＋2100 ···························································· 12分 27．（本题10分）

解：（1）B． ································································································ 2分

（2）解法一：过点B作BH垂直OC，垂足为H．

1OH1OH1

∵B在射线OA上的射影值为，∴＝，∵OB＝OA ，∴＝， 2OA2OB2∵CA＝OA，∴

OB1OHOB

＝，∴＝．又∵∠O＝∠O， OC2OBOC

B O H A C ∴△OHB∽△OBC． ················································································ 6分 ∴∠OBC＝∠OHB＝90°．∴OB⊥BC，∵点B是圆O上的一点， ∴BC是圆O的切线． ·············································································· 8分 解法二：连接AB，过点B作BH垂直OC，垂足为H．

1OH1OH1

∵B在射线OA上的射影值为，∴＝，∵OB＝OA ，∴＝＝cos∠O，

2OA2OB2∴∠O＝60°．∵OB＝OA，∴△OBA是等边三角形，∴∠OAB＝60°． ·············· 4分

∵AC＝OA，∴AB=AC，∴∠ABC=∠C，∴∠C=30°． ··································· 6分 ∴∠OBC＝90°．∴OB⊥BC，∵点B是圆O上的一点， ∴BC是圆O的切线． ·············································································· 8分 13（3）y＝0 (≤x＜)； ··············································································· 10分

24333

y＝2x－（≤x≤） ········································································· 12分

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