2024年中考押題預(yù)測(cè)卷（江蘇蘇州卷）-數(shù)學(xué)（參考答案）VIP

絕密★啟用前2024年中考押題預(yù)測(cè)卷【江蘇蘇州卷】數(shù)學(xué)一、選擇題（本大題共8小題，每小題3分，共24分，在每小題所給出的四個(gè)選項(xiàng)中，恰有一項(xiàng)是符合題目要求的，請(qǐng)將正確選項(xiàng)前的字母代號(hào)填涂在答題卡相應(yīng)位置上）．12345678CADCCADC二、填空題（本大題共8小題，每小題3分，共24分，請(qǐng)把答案填寫在答題卡相應(yīng)位置上）9． 10． 11．且 12．13．5 14．3 15．3 16．三、解答題（本大題共11小題，共82分，請(qǐng)?jiān)诖痤}卡指定區(qū)域內(nèi)作答，解答時(shí)應(yīng)寫出文字說(shuō)明、證明過(guò)程或演算步驟）17．（5分）【解析】原式．····································5分18．（5分）【解析】由得：，···································1分由得：，···································2分則不等式組的解集為，···································3分不等式組的整數(shù)解的和是．···································5分19．（6分）【解析】，···································3分即，解得：，，···································4分∵m是的一個(gè)根，且∴，∴原式．···································6分20．（6分）【解析】（1）∵，∴，∵∴···································3分（2）∵，∴∵∴···································4分∵∴∴···································6分21．（6分）【解析】（1）根據(jù)題意得，恰好選中B烹飪的概率為：，故答案為：，···································2分（2）解：列表如下：ABCDABCD由表可知，總共有12種情況，其中恰好選中B烹飪、C陶藝的情況有2種，∴好選中B烹飪、C陶藝的概率為：，故答案為：．···································6分22．（8分）【解析】（1）根據(jù)條形統(tǒng)計(jì)圖得D等級(jí)的人數(shù)有6人，根據(jù)扇形統(tǒng)計(jì)圖得D等級(jí)的百分比是，所以這次抽樣調(diào)查共抽取的人數(shù)是：(人)；···································2分（2）由（1）得這次抽樣調(diào)查共抽取的人數(shù)是60人，由扇形統(tǒng)計(jì)圖得C等級(jí)的百分比是，C等級(jí)的人數(shù)為：（人），補(bǔ)全條形統(tǒng)計(jì)圖如下：

根據(jù)扇形統(tǒng)計(jì)圖得：，故A等級(jí)的百分比為，所以A等級(jí)所在扇形圓心角的度數(shù)為：；···································4分（3）解：根據(jù)扇形統(tǒng)計(jì)圖得：，故A等級(jí)的百分比為，所以該校有1500名學(xué)生，估計(jì)該校學(xué)生答題成績(jī)?yōu)锳等級(jí)和B等級(jí)共有：（人），答：該校有1500名學(xué)生，估計(jì)該校學(xué)生答題成績(jī)?yōu)锳等級(jí)和B等級(jí)共有900人.···································8分23．（8分）【解析】（1）由題意知，，，，∵，∴，∵，即，解得，，∴土坡的水平距離為；···································3分（2）如圖，延長(zhǎng)交于，則，

由題意知，，∵，∴，···································5分由勾股定理得，，解得，，∴，∴，∴，∴，即，···································7分∴，∴樹高為．···································8分24．（8分）【解析】（1）如圖，延長(zhǎng)交軸于點(diǎn)，∵點(diǎn)A是反比例函數(shù)圖象上一點(diǎn)，過(guò)點(diǎn)A作y軸的平行線，交函數(shù)的圖象于點(diǎn)B，且∴，解得，故k的值為；···································3分（2）如圖，過(guò)點(diǎn)作，∵點(diǎn)A的橫坐標(biāo)為4，點(diǎn)A是反比例函數(shù)圖象上一點(diǎn)，∴，···································4分∵平行于y軸，∴點(diǎn)的橫坐標(biāo)為4，解得，···································5分∴正比例函數(shù)的圖象與反比例函數(shù)圖象的交點(diǎn)的坐標(biāo)為，···································6分故的面積為．···································8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，，，，，，，的平分線交于點(diǎn)，，，，···································7分過(guò)點(diǎn)作于點(diǎn)，，，···································8分，···································9分．···································10分26．（10分）【解析】（1）解：根據(jù)題意，對(duì)而言，，故點(diǎn)A是“復(fù)興點(diǎn)”；對(duì)而言，，故點(diǎn)B是“復(fù)興點(diǎn)”；對(duì)而言，，故點(diǎn)C不是“復(fù)興點(diǎn)”；故答案為：A，B；···································2分（2）解：當(dāng)時(shí)，∴，∴，∴，∴；當(dāng)時(shí)，∴，∴，∴，∴；當(dāng)時(shí)，∴，∴，∴，∴；當(dāng)時(shí)，∴，∴，∴，∴；畫圖如下：···································4分（3）解：當(dāng)時(shí)，∵反比例函數(shù)的圖像上存在4個(gè)“復(fù)興點(diǎn)”，∴反比例函數(shù)的圖像與，的圖像各有兩個(gè)交點(diǎn)，聯(lián)立方程組，，化簡(jiǎn)得，，∴，解得，∴；當(dāng)時(shí)，解：當(dāng)時(shí)，∵反比例函數(shù)的圖像上存在4個(gè)“復(fù)興點(diǎn)”，∴反比例函數(shù)的圖像與，的圖像各有兩個(gè)交點(diǎn)，聯(lián)立方程組，，化簡(jiǎn)得，，∴，解得，∴；綜上，當(dāng)或時(shí)，反比例函數(shù)的圖像上存在4個(gè)“復(fù)興點(diǎn)”；···································6分（4）解：當(dāng)時(shí)，，∴一次函數(shù)的圖像經(jīng)過(guò)定點(diǎn)，當(dāng)一次函數(shù)的圖像經(jīng)過(guò)（2）中函數(shù)圖像的點(diǎn)時(shí)，，解得；當(dāng)一次函數(shù)的圖像經(jīng)過(guò)（2）中函數(shù)圖像的點(diǎn)時(shí)，，解得；當(dāng)一次函數(shù)的圖像經(jīng)過(guò)（2）中函數(shù)圖像的點(diǎn)時(shí)，，解得；當(dāng)一次函數(shù)的圖像經(jīng)過(guò)（2）中函數(shù)圖像的點(diǎn)時(shí)，，解得，如圖，，結(jié)合函數(shù)圖像可知：當(dāng)時(shí)，復(fù)興點(diǎn)的個(gè)數(shù)為0；當(dāng)或時(shí)，復(fù)興點(diǎn)的個(gè)數(shù)為1；···································8分當(dāng)或或時(shí)，復(fù)興點(diǎn)的個(gè)數(shù)為2．···································10分27．（10分）【解析】（1）解：對(duì)于拋物線，令，可得或，，，令，可得，，，，故答案為：45；···································2分（2），，，，與都是等腰直角三角形，與的周長(zhǎng)之比為，，，，，，解得：或，當(dāng)時(shí)，位于A的左側(cè)，不符合題意，舍去，拋物線的解析式為；···································5分（3）如圖，連接，設(shè)拋物線對(duì)稱軸與x軸交于點(diǎn)E，由（2）可知，，點(diǎn)P在拋物線的對(duì)稱軸直線上，設(shè)，，是等腰直角三角形，，，解得：或，···································7分位于x軸下方，則不合題意，