2023年3月山东省济南市高新区二模检测数学卷

这是一份2023年3月山东省济南市高新区二模检测数学卷，文件包含202303高新二模-数学-试题docx、202303高新二模-数学-评分标准docx等2份试卷配套教学资源，其中试卷共12页， 欢迎下载使用。
一、选择题
二、填空题:（本大题共6个小题，每小题4分，共24分．）
11．（m+2）（m﹣2）． 12．47． 13．2+2． 14．2023． 15．y＝2x+2． 16．②③．
三、解答题:（本大题共10个小题，共86分．解答应写出文字说明、证明过程或演算步骤．）
17．（本题6分）解：原式＝1-3-2+23······························································································4分
=3-1··························································································6分
18．（本题6分）解：解第一个不等式得：x＞0········································································2分
解第二个不等式得：x≤1········································································4分
∴不等式组的正整数解是：0＜x≤1···························································5分
则整数解是：1······················································································6分
19．（本题6分）证明：∵四边形ABCD为菱形，
∴AD＝CD＝AB＝BC，∠A＝∠C···························································2分
∵BM＝BN，
∴AB﹣BM＝BC﹣BN，
即AM＝CN·······················································································4分
在△AMD和△CND中，
AM=CN∠A=∠CAD=CD，
∴△AMD≌△CND（SAS）···································································5分
∴DM＝DN·······················································································6分
20．（本题8分）解：（1）50，18··························································································2分
（2）5，6·····························································································4分
（3）150×（8×4+5×18+6×20+7×4）＝5.4（篇）·············································6分
答：本次抽查的学生平均每人阅读的篇数为5.4篇；
（4）抽查学生中阅读4篇的有8人，占抽查学生的16%，
所以1000×16%＝160（人）··································································8分
答：估计该校学生在这一周内文章阅读的篇数为4篇的人数有160人．
21．（本题8分）解：（1）∵斜坡的坡度为1：3，∴BDAB=13·····················································1分
∵BD＝CD﹣CB＝2.2（米）·····································································2分
在Rt△ABD中，AB＝3BD＝6.6（米）························································3分
故AD=2．22+6．62=11105≈7.04（米）···············································4分
答：斜面AD的长度应约为7.04米．
（2）过C作CE⊥AD，垂足为E，
∴∠DCE+∠CDE＝90°，
∵∠BAD+∠ADB＝90°，
∴∠DCE＝∠BAD，
∴tan∠BAD＝tan∠DCE=DEEC=13·······························································5分
设DE＝x米，则EC＝3x米，
在Rt△CDE中，3.22＝x2+（3x）2····························································6分
解得：x≈1.011
则3x＝3.033·························································································7分
∵3.033＞2.8，
∴货车能进入地下停车场········································································8分
22．（本题8分）（1）证明：连接OB，如图，
∵AD是⊙O的直径，
∴∠ABD＝90°······················································································1分
∴∠A+∠ADB＝90°，
∵BC为切线，
∴OB⊥BC····························································································2分
∴∠OBC＝90°，
∴∠OBA+∠CBP＝90°，
而OA＝OB，
∴∠A＝∠OBA······················································································3分
∴∠CBP＝∠ADB··················································································4分
（2）解：∵OP⊥AD，
∴∠POA＝90°，
∴∠P+∠A＝90°，
∴∠P＝∠D·························································································5分
∴△AOP∽△ABD··················································································6分
∴APAD=AOAB，即2+BP8=42·············································································7分
∴BP＝14·····························································································8分
23．（本题10分）解：（1）设每个足球的进价为x元，则每个排球的进价为（x+15）元···················1分
根据题意得3000x=3600x+15····································································································3分
解得x＝75···················································································································4分
经检验x＝75是原分式方程的解·······················································································5分
∴x+15＝75+15＝90（元）．
∴篮球的进价为75元，排球的进价为90元．
答：足球的单价为75元，排球的单价为90元······································································6分
（2）设该学校可以购进排球a个，则购进足球（100﹣a）个·················································7分
根据题意，得90a+75（100﹣a）≤8000···············································································8分
解得a≤1003·················································································································9分
∵a是整数，
∴a＝33，
答：最多可以购进排球33个··························································································10分
24．（本题10分）解：（1）将A（1，a）和B（b，2）代入y1＝-2x+8，
解得点A（1，6），B（3，2）··························································································2分
将点A（1，6）代入y2=mx，解得m＝6，即y2=6x··································································3分
（2）作B点关于y轴的对称点B'，连接AB'交y轴于点P，连接PB，
∴PB＝PB'，
∴PB+PA+AB＝PB'+AP+AB≥AB'+AB，
当A、P、B'三点共线时，△PAB的周长最小，
∵B（3，2），
∴B'（﹣3，2）·············································································································4分
设直线AB'的解析式为y＝k'x+b'，
∴-3k'+b'=2k'+b'=6，解得k'=1b'=5，
∴y＝x+5·····················································································································6分
∴P（0，5）················································································································7分
（3）D点坐标为（4，3）或（﹣2，9）或（2，1）····························································10分
25．（本题12分）解：（1）1，22··························································································2分
（2）仍然存在
证明：∵AB＝AC，2AD=AB，2AE=AC，
∴AD=AE
∵∠DAE＝∠BAC，
∴∠DAE﹣∠BAE＝∠BAC﹣∠BAE，
即∠BAD＝∠CAE·········································································································3分
在△ABD和△ACE中，AB=AC∠BAD=∠CAEAD=AE，
∴△ABD≌△ACE（SAS）·······························································································4分
∴BD＝CE，即BD：CE=1······························································································5分
∵△ABC是等腰直角三角形，AN⊥BC，
∴AB=2AN
由旋转的性质知，∠DAB＝∠MAN＝α，
∵2AD=AB，2AE=AC，
∴△ADE∽△ABC·········································································································6分
∴AMAN=ADAB=12，
∴△AMN∽△ADB········································································································7分
∴ANAB=MNDB，
∴MNBD=22，即BD=2MN······························································································8分
（3）FB的长为655或1855·······························································································12分
26．（本题12分）解：（1）由题意得：9a-3b+34=0a+b+34=0····························································2分
解得a=-14b=-12．抛物线y1所对应的函数解析式为y=-14x2-12x+34···········································3分
（2）当x＝﹣1时，y=-14+12+34=1，∴D（﹣1，1）·······················································4分
设直线AD的解析式为y＝kx+b，
∴-3k+b=0-k+b=1，解得k=12b=32，
∴直线AD的解析式为y=12x+32······················································································5分
如答图1，当M点在x轴上方时，
∵∠M1CB＝∠DAC，
∴DA∥CM1，
设直线CM1的解析式为y=12x+b1···················································································6分
∵直线经过点C，
∴-12+b1=0，解得：b1=12··························································································7分
∴直线CM1的解析式为y=12x+12，
∴y=12x+12y=-14x2-12x+34，解得：x=-2+5，x=-2-5（舍去），∴m=-2+5··················8分
（3）P点坐标为(0，34)或P(2，-54)···············································································12分题号
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答案
A
C
A
B
A
A
C
A
B
A