2021至2022学年第一学期期末学业水平测试 高新初中数学九年级参考答案及评分标准
一、选择题 题号 答案 1 C 2 C 3 B 4 A 5 C 6 D 7 A 8 B 9 C 10 C 11 A 12 D 二、填空题:（本大题共6个小题，每小题4分，共24分．） 13．﹣1 14．∠ABD＝∠C或∠ADB＝∠ABC等． 15．y＝（x+1）2﹣2或y＝x2+2x-1． 16．3√5． 17．114． 18．①③④．
三、解答题:（本大题共12个小题，共78分．解答应写出文字说明、证明过程或演算步骤．）
19．（本题6分）解：原式＝+1+1﹣1·················································································4分
21
=·····························································································6分
2
3
20．（本题6分）解：由题意得，设入y＝a（x+1）2﹣4····························································2分
代入（0，﹣3）得a＝1··········································································4分 ∴y＝x2+2x﹣3······················································································6分
21．（本题6分）证明：∵∠1＝∠2，∠DPA＝∠CPB································································2分
∴△ADP∽△BCP(AA) ·····································································3分 ∴
PAPD····················································································5分 PBPC∵PB=3，PC=1，PD=2
∴PA=6···························································································6分
22．（本题8分）解：（1）柱子OA的高度为4米·······································································2分 （2）在y＝﹣x2+2x+中，
47
7
当y＝0时﹣x2+2x+=0··································································································4分
4
7
∴x1=
√11√11+1，x2＝1−·································································································622
分
又∵x＞0， ∴解得x=
√11+12
米·······································································································7分
√11+12
答：水池的半径至少要米才能使喷出的水流不至于落在池外··········································8分
23．（本题8分）（1）证明：如图，连接OD············································································1分 ∵DE与⊙O相切于点D，
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∴DE⊥OD··················································································································2分 ∴∠ODE＝90°， ∵OD＝OA， ∴∠ODA＝∠OAD ∵AD平分∠BAF， ∴∠OAD＝∠DAF，
∴∠ODA＝∠DAF·········································································································3分 ∴OD∥AF，
∴∠AED＝180°﹣∠ODE＝90°，
∴DE⊥AF··················································································································4分 （2）如图，连接BD·····································································································5分 ∵AB是⊙O的直径，
∴∠ADB＝90°·············································································································6分 ∴∠AED＝∠ADB， ∵∠EAD＝∠DAB，
∴△AED∽△ADB·········································································································7分 ∴𝐴𝐷=𝐴𝐵， ∵AE＝8，AB＝10，
∴AD=√𝐴𝐸⋅𝐴𝐵=√8×10=4√5···················································································8分 24．（本题10分）解：（1）1.5·························································································2分 （2）结合光的反射原理得：∠CED＝∠AEB·······································································3分 在Rt△CED和Rt△AEB中，
∵∠CDE＝∠ABE＝90°，∠CED＝∠AEB，
∴Rt△CED∽Rt△AEB···································································································5分 ∴
𝐷𝐶𝐴𝐵1.5𝐴𝐸
𝐴𝐷
=
𝐷𝐸𝐸𝐵2
····················································································································6分
即𝐴𝐵=20，
解得AB＝15（m）·········································································································7分 答：铁塔AB的高度是15m·····························································································8分 （3）受天气条件影响，没有太阳光线，或旗杆底部不可能达到相等·······································10分
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25．（本题10分）解：（1）∵点A（1，3）在反比例函数y=𝑥的图象上，则k＝3···························1分 ∴反比例函数的解析式为y=·························································································2分
𝑥3
𝑘
∵点B（3，n）在反比例函数y=的图象上，则n＝1···························································3分
𝑥
3
（2）①由（1）知，n＝1，∴B（3，1）， 设直线AB的解析式为y＝ax+b，
𝑎=−1代入点A（1，3），B（3，1），得{，
𝑏=4
∴直线AB的解析式为y＝﹣x+4·······················································································4分 ∴D（0，4），则OD＝4，
令y＝0，则﹣x+4＝0，则C（4，0）·················································································5分 ∴OC＝OD， ∵∠COD＝90°，
∴∠OCD＝∠ODC＝45°， 由折叠知，∠OCD＝∠ECD＝45°， ∴∠OCE＝90°， ∴CE⊥x轴， ∴点F的横坐标为4， ∴y=4，
∴F（4，4）·················································································································6分 ②存在
假设存在，设P（m，0），由①知，F（4，4），D（0，4），
∴PF2＝（m﹣4）2+（4）2，PD2＝m2+42，DF2＝42+（4−4）2·················································7分 ∵△DPF是以DF为斜边的直角三角形， ∴DF2＝PF2+PD2，
∴42+（4−4）2＝（m﹣4）2+（4）2+m2+42··········································································8分 ∴m2﹣4m+3＝0 ∴m＝1或m＝3，
即在x轴上是存在点P，点P（1，0）或（3，0），使得△DPF是以DF为斜边的直角三角形·······10分 26．（本题12分）解：（1）①BE＝2CD··················································································2分
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3
3
3
3
3
33
②仍然成立··················································································································3分 理由：∵∠ACB＝∠ADE＝90°，ABC＝∠AED＝30°
∴△ACB∽△ADE(AA)，且∴∠CAD＝∠BAE，
AC1········································································5分 ·
AB2ACAB ADAE∴△ACD∽△ABE(SAS) ························································································7分 ∴
ACCD1 ABBE2∴BE＝2CD········································································································8分 （2）综上所述，线段CD的长为2√10或4√10···································································10分
27．（本题12分）解：（1）由已知，得B（3，0），C（0，3）， 3=𝑐𝑏=−4∴{，解得{， 0=9+3𝑏+𝑐𝑐=3
∴抛物线解析式为y＝x2﹣4x+3························································································2分 y＝x2﹣4x+3＝（x﹣2）2﹣1，顶点坐标为P（2，﹣1）·······················································3分 （2）当0＜x＜3时，在此抛物线上任取一点E，连接CE、BE，经过点E作x轴的垂线FE，交直线BC于点F，
设点F（x，﹣x+3），点E（x，x2﹣4x+3），
∴EF＝﹣x2+3x·············································································································4分 ∴S△CBE＝S△CEF+S△BEF=2EF•OB， =−x2+x，
23
23
9
1
=−2（x−2）2+8， ∵a=−2＜0，
∴当x=2时，S△CBE有最大值···························································································5分 ∴y＝x2﹣4x+3=−4，
∴E（2，−4）···············································································································6分
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3
3
3
33
327
（3）①由（1）得A（1，0）， 连接BP，
∵∠CBA＝∠ABP＝45°··································································································7分 ∴当𝐵𝑃=𝐵𝐴时，△ABC∽△PBN， ∴BN＝3．
∴N1（0，0）··············································································································8分 ∴当𝐵𝑃=𝐵𝐶时，△ABC∽△NBP， ∴BN=3．
∴N′（，0）················································································································9分
372𝐵𝑁
𝐵𝐴
𝐵𝑁
𝐵𝐶
②（2，2）；（2，-4）；（2，4）························································································12分
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