數(shù)學（浙江卷）（參考答案及評分標準）

2024年中考第一次模擬考試（浙江卷）數(shù)學·參考答案第Ⅰ卷一、選擇題（本大題共10個小題，每小題3分，共30分．在每個小題給出的四個選項中，只有一項符合題目要求，請選出并在答題卡上將該項涂黑）12345678910DBCBDCABBA第Ⅱ卷二、填空題（本大題共6小題，每小題3分，共18分）11．212．7213．14．43三、解答題（本大題共8個小題，共72分．解答應寫出文字說明，證明過程或演算步驟）17．【答案】（1）；（2）【分析】本題考查了實數(shù)的運算以及解一元一次不等式；（1）分別根據(jù)零指數(shù)冪的定義，絕對值的性質以及二次根式的性質，計算即可；（2）不等式去括號，移項，合并同類項，化系數(shù)為1即可．【詳解】（1）原式；·························································3分（2），去括號，得，移項，得，合并同類項，得．························································6分18．【答案】錯誤步驟的序號為①，解法見詳解．【分析】本題考查檢查解分式方程；錯誤步驟的序號為①，解方程去分母轉化為整式方程，，進而解這個整式方程，最后檢驗，即可求解．【詳解】解：錯誤步驟的序號為①，························································1分去分母得：去括號得：移項得：…③，合并同類項得：…④，························································3分檢驗：當時，，························································5分∴是原分式方程的解．························································6分19．【答案】(1)見解析(2)，(3)二，理由見解析【分析】本題考查統(tǒng)計圖分析，涉及中位數(shù)、加權平均數(shù)、眾數(shù)，（1）根據(jù)這30名學生第一次競賽成績和第二次競賽成績得分情況統(tǒng)計圖可得橫坐標是89，縱坐標是90的點即代表小松同學的點；（2）根據(jù)平均數(shù)和中位數(shù)的定義可得m和n的值；（3）根據(jù)平均數(shù)，眾數(shù)和中位數(shù)進行決策即可．【詳解】（1）解：（1）如圖所示．·······························2分（2），∵第二次競賽獲卓越獎的學生有16人，成績從小到大排列為：90

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98，∴第一和第二個數(shù)是30名學生成績中第15和第16個數(shù)，∴，∴，；························································6分（3）可以推斷出第二次競賽中初三年級全體學生的成績水平較高，理由是：第二次競賽學生成績的平均數(shù)、中位數(shù)、眾數(shù)都高于第一次競賽．答：二，第二次競賽學生成績的平均數(shù)、中位數(shù)、眾數(shù)都高于第一次競賽．·············8分20．【答案】任務1：剪掉的正方形的邊長為．任務2：當剪掉的正方形的邊長為時，長方形盒子的側面積最大為．【分析】此題主要考查了一元二次方程和二次函數(shù)的應用，找到關鍵描述語，找到等量關系準確地列出方程和函數(shù)關系式是解決問題的關鍵．任務1：假設剪掉的正方形的邊長為，根據(jù)長方形盒子的底面積為，得方程，解所列方程并檢驗可得；任務2：側面積有最大值，設剪掉的正方形邊長為，盒子的側面積為，利用長方形盒子的側面積為：得出即可．【詳解】解：任務1：設剪掉的正方形的邊長為，則，即，解得（不合題意，舍去），，答：剪掉的正方形的邊長為．························································3分任務2：側面積有最大值．理由如下：設剪掉的小正方形的邊長為，盒子的側面積為，則與的函數(shù)關系為：，即，即，························································6分∴時，．························································8分即當剪掉的正方形的邊長為時，長方形盒子的側面積最大為．21．【答案】(1)支點C離桌面l的高度；(2)面板上端E離桌面l的高度是增加了，增加了約【分析】（1）作，先在求出的長，再計算即可得答案；（2）分別求出時和時，的長，相減即可．【詳解】（1）解：如下圖，作，，，，，支點C離桌面l的高度；···············································4分（2），，當時，，························································5分當時，，························································6分，面板上端E離桌面l的高度是增加了，增加了約．···································10分【點睛】本題考查了解直角三角形的應用，解題的關鍵是作輔助線，構造直角三角形．22．【答案】(1)(2)(3)【分析】本題考查了相似三角形的判定與性質、正方形的性質等知識點，掌握相似三角形判定定理的內容是解題關鍵．（1）證可得，結合即可求解；（2）由可得，進一步可得，據(jù)此即可求解；（3）由（1）可得，證得即可求解．【詳解】（1）解：由題意得：∴∴即：解得：························································2分（2）解：∵，∴∴························································3分由（1）可得：∴∴∵，∴························································5分解得：························································6分（3）解：由（1）得：即：解得：························································7分∵，∴∴即：∴整理得：························································8分∵∴，又∴故：························································10分23．【答案】(1)120(2)2(3)(4)見解析，【分析】本題主要考查了垂徑定理在圓中的應用，最后一問由“共頂點，等線段”聯(lián)想到旋轉，是此題的突破口，同時，要注意頂角為的等腰三角形腰和底邊比是固定值．（1）由已知得到垂直平分，故得到，證明為等邊三角形即可得到答案；（2）由于直徑，根據(jù)垂徑定理可以得到是的中點，要求最大值即求最大值，當為直徑時，有最大值，即可得到答案；（3）根據(jù)垂徑定理得到，證明，由（1）得，即可得到答案；（4）將繞A點順時針旋轉至，得到，證明，過A作于G，則，根據(jù)勾股定理證明．【詳解】（1）解：連接，，、，，，，，，，，的度數(shù)為；························································2分（2）解：由題可知，為直徑，且，由垂徑定理可得，，連接，是的中點，，當三點共線時，此時取得最大值，且，的最大值為；························································4分（3）解：連接，，，，平分，，，，，，，；

························································6分（4）證明：由題可得，直徑，垂直平分，如圖4，連接，，則，由（1）得，將繞A點順時針旋轉至，，，，四邊形為圓內接四邊形，，，、D、P三點共線，，························································7分過A作于G，則，，在中，，設，則，，，························································8分，，························································10分為定值．························································12分24．【答案】(1)；(2)①，；②的最小值為．【分析】（1）將點、的坐標代入拋物線，利用待定系數(shù)法求得解析式；（2）①由坐標求出解析式，然后根據(jù)四邊形是平行四邊形和得出，再分類討論求得和的坐標；②求出解析式，交點為，再求出坐標，然后由兩點間距離公式求出和長度，因為旋轉不改變長度，所以長度不變，當旋轉到軸上時，此時最短，所以此時等于，然后帶入計算即可．【詳解】（1）解：①∵拋物線交軸于點和點，∴將、坐標代入有，解得∴拋物線的表達式為；························································2分（2）解：∵拋物線的表達式為，∴，設直線的解析式為∵，，∴解得∴直線的解析式為························································3分∵為與軸交點，∴，∴，∵四邊形是平行四邊形∴且，且點在點下方，∵且在軸上∴，∵，∴，或，························································4分若為，，∵，故，若為，，∵，此時，矛盾，舍去綜上，；························································6分②最小值為如圖，設的解析式為∵拋物線交軸于點，∴點的坐標為，將點，、，的坐標代入得解得∴的解析式為與相交于點∴解得所以點的坐標為························································8分設直線的解析式為將點、的坐標代入直線的解析式得解得所以直線的解析式為·····································