所以，不等式组的解集是－2＜x≤1． ··························· 5分 画图正确（略）． ······················································ 7分 19．（本题7分）
（1）126； ···································································· 2分 （2）图略； ··································································· 4分 （3）在抽取的样本中，“比较喜欢”数学的人数所占的百分比为
1－32%－10%－23%＝35%， ········································ 5分 由此可估计，该校1000名学生中，“比较喜欢”数学的人数所占的百分比35%，
1000×35%＝350（人）． ············································ 6分 答：估计这些学生中，“比较喜欢”数学的人数约有350人． 7分
20．（本小题满分8分）
证明：（1）∵ 四边形ABCD是平行四边形，∴ AB//CD，AB＝DC．∴ ∠ABC＝∠DCE．
∵ AC//DE，∴ ∠ACB＝∠DEC． ·································· 3分
在△ABC和△DCE中，∠ABC＝∠DCE，∠ACB＝∠DEC ，AB＝DC．
∴△ABC≌△DCE（AAS）． ··································· 4分 （2）由（1）知△ABC≌△DCE，则有BC＝CE． ∵ CD＝CE， ∴ BC＝CD．
∴四边形ABCD为菱形． ············································· 7分
∴AC⊥BD． ···························································· 8分 21．（本题7分）
列表或树状图表示正确； ············································· 3分 ∵共有8种等可能的结果，
通过一次“手心手背”游戏， 小明先跳绳的有2种情况 ······ 5分 2 1 ∴通过一次“手心手背”游戏，小明先跳绳的概率是：8 ＝ 4． 1 答：通过一次“手心手背”游戏，小明先跳绳的概率是4． ···· 7分 22．（本题6分）
方法1： 方法2：
··················································································· 6分 23．（本题7分）
解：过点A作AD⊥OB于点D.
由题意得AN⊥MN，OB⊥MN，AD⊥OB，∴四边形ANMD是矩O 形，
A D B C N M ∴DM＝AN， ······························································ 2分 设OB＝OA＝x cm，在Rt?OAD中，∠ODA＝90°，
x＋5－14OD
cos∠AOD＝OA ＝ ≈0.6． ······························· 5分 x解得x＝15cm．
经检验，x＝15为原方程的解．
答：细线OB的长度是15cm． ······································· 7分
24．（本小题满分7分）
解：设每千克樱桃应降价x元，根据题意，得 ························ 1分
（60－x－40）（100＋10x） ＝ 2240． ·························· 4分 解得：x1＝4，x2＝6． ·················································· 6分 答：每千克樱桃应降价4元或6元． ······························ 7分
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
－8x＝3， ··············································· 4分 x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． B 1OH1OH∵B在射线OA上的射影值为2，∴OA＝2，∵OB＝OA ，∴H ＝O OBA 12， OB1OHOB ∵CA＝OA，∴OC＝2，∴OB＝OC．又∵∠O＝∠O， ∴△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． 1OH1OH∵B在射线OA上的射影值为2，∴OA＝2，∵OB＝OA ，∴OB＝1 2＝cos∠O， ∴∠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 (2≤x＜4)； ················································ 10分 333 y＝2x－2（4≤x≤2） ············································· 12分 C
2020年中考数学全真模拟试卷及答案(共五套)
