絕密★啟用前2024年中考押題預測卷【江蘇無錫卷】數學一、選擇題（本大題共10小題，每小題3分，共30分，在每小題所給出的四個選項中，恰有一項是符合題目要求的，請將正確選項前的字母代號填涂在答題卡相應位置上）．12345678910ADBDACDAAA二、填空題（本大題共8小題，每小題3分，共24分，請把答案填寫在答題卡相應位置上）11． 12． 13．1 14．215．B 16 17． 18．//三、解答題（本大題共10小題，共96分，請在答題卡指定區域內作答，解答時應寫出文字說明、證明過程或演算步驟）19．（9分）【解析】解：（1）；················································4分（2）···············································9分20．（9分）【解析】（1），，，，∴，∴，；···············································4分（2），解不等式①，得，解不等式②，得，∴不等式組的解集為．···············································9分21．（9分）【解析】（1）證明：如圖，連接交于點O，∵四邊形是平行四邊形，∴，∵，∴，∴，∵，∴四邊形BEDF是平行四邊形；···············································4分（2）解：∵，∴，∵，，∴···············································9分22．（9分）【解析】（1）用樹狀圖表示為：由圖可知，共有4種等可能結果，其中P、Q間沒有電流通過的只有1種，有電流通過的有3種，∴之間電流通過的概率是；····································5分（2）畫樹狀圖得：由圖可知，共有8種等可能結果，其中沒有電流通過的只有1種，有電流通過的有7種，∴之間電流通過的概率是．故答案為：．····································9分23．（10分）【解析】（1）由題意得，，．故答案為：5；····································2分（2）解：統計圖中B組對應扇形的圓心角為，故答案為：144；····································4分（3）解：由題意可知，閱讀時間在范圍內的數據的眾數是45，調查的20名同學課外閱讀時間的中位數是．故答案為：45，；····································6分（4）解：（人），答：估計全校800名同學課外閱讀時間不少于的人數大約為480人．··································10分24．（10分）【解析】（1）如圖所示，過點M作的垂線，交于O，以點O為圓心，為半徑畫圓，則圓O即為所求；····································3分如圖所示，過點O作于N，∵平分，，∴，∴射線與相切；····································6分

（2）解：∵和為的切線，∴，，，∴，∴和都是等腰直角三角形，∴，即，∴的劣弧與所圍成圖形的面積．故答案為：．····································10分25．（10分）【解析】（1）證明：連接，如圖，是的平分線，，，為的直徑，，，，，為的半徑，直線是的切線；····································4分（2）解：為的直徑，，，，，，，····································6分的平分線交于點，，，，····································8分過點作于點，，，，．····································10分26．（10分）【解析】（1）設與的函數表達式為，把和分別代入得：，解得：，∴與的函數表達式為；····································4分（2）解：當時，，····································5分∵，∴隨的增大而減小，∴當時，；····································6分當時，，····································7分∵不在范圍內，當時，隨的增大而減小，∴當時，；····································9分綜上述，第天時，當天的銷售利潤最大，最大銷售利潤是元．····································10分27．（10分）【解析】（1）①四邊形是矩形，，，由折疊知，，，，，在和中，，，，，設，則，，在中，由勾股定理得：，即，解得：，則；····································2分②如圖，連接交于點，過點作于點，，（對頂角），，，，，則，，（對頂角），，，，，，，，；····································6分（2）當落在直線上面時，如圖，過作于，，，，，又，，由翻折可知，在中，，，又，在中，，此時只要，點在邊上，；當落在直線下面時，如圖，過作于，同理可得，，在中，，，，，，，在中，，此時要在邊上，則即可，即，綜上，．····································10分28．（10分）【解析】（1）由題可得：，，當時，總有，，整理得：，，，，；····································2分（2）解：①注意到拋物線最大值和開口大小不變，只影響圖象左右平移，下面考慮滿足題意的兩種臨界情況：當拋物線過點時，如圖1所示，，此時，，，解得或（舍去）；當拋物線過點時，如圖2所示，，此時，，，解得：或（舍去），綜上所述，；····································4分②同①考慮滿足題意的臨界情形：當拋物線過點時，如圖3所示，，此時，，，解得：或（舍去），·······················5分當拋物線過點時，如圖4所示，，此時，，，解得或（舍去），··················