钦州市2023年学科素养测试-数学-试卷

2023年钦州市初中学科素养测试参考答案

数  学

一、选择题（本大题共12小题，每小题3分，共36分）

二、填空题（本大题共6小题，每小题2分，共12分）

13．x≥－6； 14．（x+1）（x-1）； 15．（-2，-3）；  16．52.5；  17．150；  18．．

三、解答题（本大题共72分）

19．解：原式 ························································3分

             ·························································4分

       .     ······················································6分

20．解：

①式等号两边同时乘以2得：2x－4y＝8③，···························1分

③－②得：-y＝3，      ········································2分

解得：y＝-3，·················································3分

把y＝-3带入①式得：x－2×（-3）＝4，      ·························4分

解得x＝-2.···················································5分

所以方程组的解为·························6分

21．解：（1）△A1B1C1如图所示.·····5分

（2）AE为所作图形．·······10分

22．证明：（1）∵四边形ABCD是平行四边形，

∴AB＝CD，∠B＝∠D．··································2分

在△ABE和△CDF中，

····································3分

∴△ABE≌△CDF（SAS）．·······························4分

（2）∵四边形ABCD是平行四边形，

∴AD＝BC，且AD∥BC．··································6分

∵BE＝DF，

∴AF＝EC，且AF∥EC．··································8分

∴四边形AECF是平行四边形．······························9分

又∵∠AEC＝90°，

∴四边形AECF为矩形．··································10分

23．解：（1）a＝45，b＝5，c＝44，d＝47··································4分

（2）答：两组的平均数相同，乙组的中位数、众数都比甲组的高，

所以乙组学习比较积极．··································7分

（3）根据题意得： （人）

答：估计该日登录“学习强国”学习获得积分不低于45分的人数是72人.

·······················································10分

24．解：（1）设B车间每天生产个装饰品，A车间每天生产1.5个装饰品.············1分

依题意，得

     ·····················································2分

解得，··················································3分

经检验，是分式方程的解，··································4分

则A车间每天生产720×1.5＝1080（个）

答：A车间每天生产1080个装饰品，B车间每天生产720个装饰品.······5分

（2）设安排A车间工作m天，B车间工作n天.

根据题意得：1080m＋720n＝21600，···························6分

整理得：，

成本：··················································7分

   整理得：，

因为3240＞0，所以，w随m的增大而增大，······················8分

又因为天数为正整数，所以，m为2的倍数，m取最小值为2，此时n = 27，

成本w最低，·············································9分

最低成本为：元.

答：安排A车间工作2天，B车间工作27天时，成本最低，最低成本是265680元

.       ·························································10分

25．（1）证明： 如图，连接OD，··········································1分

∵AB＝BC，OC＝OD，

∴∠BAC＝∠BCA，∠BCA＝∠ODC.·······························2分

∴∠ODC＝∠BAC，∴OD∥AB，·································3分

∵DE⊥AB，∴DE⊥OD.·········································4分

∵OD是⊙O的半径，

∴DE是⊙O的切线.···········································5分

（2）解：如图，连接BD，过点D作DH⊥BC于点H，

设BD＝k，∵，

∴CD＝2k，

∴，∴，

∴，，·····················································6分

∵，

∴，

∵，

∴，·······················································7分

∵DE⊥OD，DH⊥OE，

∴∠ODE＝∠OHD＝90°，

∵∠EOD＝∠DOH，

∴△ODE∽△OHD，∴，

∴，

∴，∴，··················································8分

由（1）知OD∥AB，

∴△BFE∽△ODE，··········································9分

∴.·····················································10分

26．解：（1）∵抛物线与x轴交于点M、N，且当时，

解得x1＝－1，x2＝3，·····································1分

∴M(－1，0)，N(3，0)；······································2分

将点M(－1，0)，B(0，1)代入抛物线y＝ax2－2ax＋c，

得解得

∴抛物线C2的解析式为；····································3分

（2）设P(t，)(－1＜t＜3)，则Q(t，)，·································4分

∴，，

························································5分

∴，∴的值为定值，该定值为2；·······························6分

（3）存在．···················································7分

由抛物线C1：可得点A(0，3)，两条抛物线的对称轴均为直线

.

∵点D是点B关于抛物线对称轴的对称点，B(0，1)，

∴D(2，1)，

如解图，连接BD，

∵AB＝BD＝2，

∴△ABD为等腰直角三角形，

∴，·····················································8分

假设存在，设点F(m，0)，分两种情况讨论：

①当AD＝FD时，，

如解图①，过点D作DC⊥x轴于点C，连接AF，DF，

则CD＝1，CF＝2－m，

由勾股定理可知，

∴，解得：，，

∴F1(，0)，F2(，0)；·········································9分

②当AD＝AF时，，

如解图②，由勾股定理可得，

∴，此方程无解，∴此种情况不存在．

综上所述，在x轴上存在点F，使得△ADF是以AD为腰的等腰三角形，点F的坐标为

(，0)或(，0)．··················································10分