2018年郴州市中考数学模拟试题答案.docx

‎2018年郴州市初中毕业学业考试模拟试卷 数学参考答案及评分标准 说明：‎ 1. 如果考生的解法与本答案的解法不同，可参照本答案的评分意见给分.‎ 2. 评卷中，不要因解答中出现错误而中断对该题的评阅，当解答中某一步出错影响了后继部分，但该步以后的解未改变这道题的内容和难度，在未发生新的错误前，可视影响的程度决定后面部分的记分，但不应超过后面部分应给分数的一半，如有严重错误，就不记分.‎ 3. 各题解答中右端所给分数，表示考生正确做到这步应得的累加分数.‎ 一、选择题（共8小题，每小题3分，共24分）‎ ‎1—5 BCADD 6—8 CBA 二、填空题（共8小题，每小题3分，共24分）‎ ‎9. ‎ ‎10. 且 ‎11. ‎ ‎12. 乙 ‎13. ‎ ‎14. 45°‎ ‎15. ‎ ‎16. 7‎ 三、解答题（17～19题每题6分，20～23题每题8分，24～25题每题10分，26题12分，共82分）‎ ‎17. 解：原式····································4分（每一知识点1分）‎ ‎·······························································5分 ‎·············································································2分 数学参考答案及评分标准 第 8 页（共 8 页）‎ 18. 解：原式·····················································2分 ‎·······································································3分 ‎ ················································································4分 ‎ ················································································5分 当时，原式··························································6分 19. 证明：如图，∵ 四边形ABCD是平行四边形，‎ ‎ ∴ AB∥DC，AB＝DC．………………………… 1分 ‎∵ DE＝AB，‎ ‎∴ DE＝DC．‎ ‎ ∴ ∠DCE＝∠DEC．…………………………3分 ‎∵ AB∥DC，‎ ‎∴ ∠ABC＝∠DCE． ‎ ‎ ∴ ∠ABC＝∠DEC． …………………………4分 又∵ AB＝DE，BE＝EB，‎ ‎∴ △ABE≌△DEB． …………………………5分 ‎∴ AE＝BD．…………………………6分 20. 解：（1）200·······················································································2分 ‎（2）图1和图2补充如下：‎ 数学参考答案及评分标准 第 8 页（共 8 页）‎ ‎30‎ ‎60‎ ‎·····················4分 小说所对应的圆心角度数为：360°×20％＝72°···········································6分 ‎（3）800×40％＝320（人）······································································7分 答：全校学生中最喜欢漫画人数约为320人。···············································8分 18. 解：（1）设一件A商品的进价为元，‎ 由题意可得：‎ 解得：‎ 经检验，是原方程的根 此时，‎ 所以一件A型商品的进价为160元，B型商品的进价为150元.·························2分 ‎（2）设A型商品m件，B型商品（250－m）件，则 ‎ ‎ 解得80≤m≤125，‎ 函数关系式为：y＝10m＋17500（80≤m≤125）；·······························5分 ‎（3）y＝10m＋17500－ma＝（10－a）m＋17500，‎ 当0＜a＜10时，y随m的增大而增大，当m＝125时利润最大，‎ ymax＝1250－125a＋17500＝18750－125a；‎ 数学参考答案及评分标准 第 8 页（共 8 页）‎ 当a＝10时，y＝17500，ymax＝17500；‎ 当a＞10时，y随m的增大而减小，‎ 当m＝80时，利润最大，ymax＝800－80a＋17500＝18300－80a．···········8分 18. ‎1分 ‎4分 ‎3分 ‎6分 ‎7分 ‎8分 ‎5分 ‎2分 19. 数学参考答案及评分标准 第 8 页（共 8 页）‎ ‎8分 ‎7分 ‎6分 18. 解：（1）3；·····················································································2分 （2） ‎∵＜x－1＞＝3，‎ ‎∴2.5≤x－1＜3.5，····················································································4分 ‎∴3.5≤x＜4.5.··························································································5分 ‎（2）解不等式组得：﹣1≤x＜＜a＞，·························································7分 由不等式组整数解恰有3个得，1＜＜a＞≤2，········································9分 故1.5≤a＜2.5；···············································································10分 19. 数学参考答案及评分标准 第 8 页（共 8 页）‎ 18. ‎（1）由四边形ABCD为菱形，可知，‎ ‎ ∴在Rt△AOB中， （3分）‎ ‎（2）在菱形ABCD中，AB∥CD，∠ADB=∠CDB 又∵四边形APFD为平行四边形 ‎∴DF=AP=t 又∵EF⊥BD于Q，且∠ADB=∠CDB ‎∴∠DEF=∠DFE ‎∴DE=DF=t ‎∴AE=10-t （4分）‎ 数学参考答案及评分标准 第 8 页（共 8 页）‎ ‎∵PE∥BD ∴△APE∽△ABD ‎∴ 即 ∴t=5‎ 故当t=5时，PE∥BD； （6分）‎ ‎（3）①∵∠FDQ=∠CDO，∠FQD=∠COD=90º ‎∴△DFQ∽△DCO ‎∴ 即 ∴QF=‎ ‎∴EF=2QF=‎ 同理，QD= （7分）‎ 过点C作CG⊥AB于点G 即 ‎∴CG=‎ ‎ （8分）‎ ‎∴‎ ‎∴， （9分）‎ ‎②当 即 （10分）‎ ‎ ‎ 数学参考答案及评分标准 第 8 页（共 8 页）‎ ‎ >10（不合题意，舍去） （11分）‎ 故存在t=4s，使得S四边形APFE=S菱形ABCD． （12分）‎ 数学参考答案及评分标准 第 8 页（共 8 页）‎