瀘州市高2020級第二次教學質量診斷性考試數(shù)  學（理科）參考答案及評分意見評分說明：1．本解答給出了一種或幾種解法供參考，如果考生的解法與本解答不同，可根據(jù)試題的主要考查內(nèi)容比照評分參考制訂相應的評分細則．2．對計算題，當考生的解答在某一步出現(xiàn)錯誤時，如果后繼部分的解答未改變該題的內(nèi)容和難度．可視影響的程度決定后繼部分的給分，但不得超過該部分正確解答應得分數(shù)的一半；如果后繼部分的解答有較嚴重的錯誤，就不再給分．3．解答右側所注分數(shù)，表示考生正確做到這一步應得的累加分數(shù)．4．只給整數(shù)分數(shù)，選擇題和填空題不給中間分．一、選擇題：題號123456789101112答案AADCCDCBBBCC 二、填空題： 13．1；     14．中的任意一個值；   15．；      16．．三、解答題：17．解：（Ⅰ）因為，  ①所以當時，，   ②············································1分由①，②相減得：，··········································· 2分即，······················································· 3分在中令得，，即，············································· 4分所以數(shù)列是以為首項，公比為的等比數(shù)列，··························· 5分所以；·····················································6分（Ⅱ）若選①．因為··························································7分··························································8分所以······················································10分．·······················································12分若選②．···················································7分··························································8分所以······················································11分．·······················································12分18．解：（Ⅰ）設“該考生報考甲大學恰好通過一門筆試科目”為事件A，“該考生報考乙大學恰好通過一門筆試科目”為事件，根據(jù)題意得：，·······················································2分························································3分；·······················································4分（Ⅱ）設該考生報考甲大學通過的科目數(shù)為，報考乙大學通過的科目數(shù)為，根據(jù)題意可知，，所以，，······································5分，························································6分，························································7分，························································8分．························································9分則隨機變量的分布列為：0123，·······················································10分若該考生更希望通過乙大學的筆試時，有，··························11分所以，又因為，所以，所以m的取值范圍是．···························12分19．證明：（Ⅰ）分別延長B1D，BA，設，連接CE，····················1分則CE即為平面與平面的交線，······················2分因為，取中點F，連接DF，························3分所以，平面，因為平面平面，且交線為，所以平面，·····················································4分因為D為棱的中點，，所以D為的中點，所以，············································5分所以平面；·····················································6分方法一：（Ⅱ）由（Ⅰ）知，因為，，所以， 在平面內(nèi)過點C作，垂足為G，則平面，·································7分分別以CB，CE，CG所在直線為x，y，z軸，建立如圖所示的空間直角坐標系，設，則，，，···················································8分則，，，，·····················································9分設平面的法向量為，則，取，······················································10分設平面的法向量為，則，取，······················································11分所以，即二面角的余弦值為．············································12分方法二：連接BF，因為四邊形為菱形，且，所以，·····································7分平面，因為平面平面，且交線為，所以平面，··································8分過點F作，連接，所以，故為二面角的平面角，·············································9分在中，，，，所以，························································10分在中，，所以，·················································11分所以，即二面角的余弦值為．·······································12分20．解：（Ⅰ）因為在C上，所以，·············································1分因為C的左焦點，所以，············································2分所以，，的方程為；··················································4分（Ⅱ）①當直線與x軸重合時，點，，，，，，所以，··················································5分②當直線與x軸不重合時，設直線的方程為，代入消去x得，因為直線與C交于點，，所以，···································6分因為，·····················································7分所以，·····················································8分（1）當m≠0時，同理可得，········································9分，·······················································10分因為，所以的取值范圍是，··········································11分（2）當時，，綜上知的取值范圍是.··········································12分21．解：（Ⅰ），·························································1分因為是函數(shù)的一個極值點，所以，得，··················································2分所以，因此在上單減，在上單增，······································3分所以當時，有最小值；·········································4分方法一：（Ⅱ）因為，所以，則在上單增，···········································5分記，當時，，當時，，則，·······················································6分記，當時，；當時，；7分所以存在唯一的，使得，當時，；當時，，所以函數(shù)在上單減，在上單增，···································8分若函數(shù)有兩個零點，只需，即，又，即，···················································9分則，設，則為增函數(shù)，，所以當時，，則，即，··················································10分令，，則在上單增，由得，··········································11分所以，所以a的取值范圍是．·········································12分方法二：（Ⅱ）若有兩個零點，即有兩個解，即有兩個解，········································5分利用同構式，設函數(shù)，·········································6分問題等價于方程有兩個解，······································7分恒成立，即單調(diào)遞增，所以，問題等價于方程有兩個解，······································8分即有兩個解，設，，即有兩個解，令，問題轉化為函數(shù)有兩個零點，·································9分因為，當時，，當時，，則在上遞增，在上遞減，·······································10分為了使有兩個零點，只需，解得，即，解得，············································11分由于，所以在和內(nèi)各有一個零點．綜上知a的取值范圍是．········································12分22．解：（Ⅰ）由，得，·····················································1分所以，·····················································2分又，，·····················································3分所以，·····················································4分即的直角坐標方程為；··········································5分（Ⅱ）曲線的普通方程為：，·········································6分直線的參數(shù)方程為：為參數(shù)），···································7分代入整理得：，··············································8分設A，B兩點所對應的參數(shù)分別為，，則，因為，所以，即或，·············································9分因為，或，滿足，所以或．····················································10分23．解：（Ⅰ）因為，························································1分若對，恒成立，則，···········································2分所以，或，··················································4分所以實數(shù)m的取值范圍是；······································5分（Ⅱ）由（Ⅰ）知，的最小值為，所以，································6分所以或，因為，所以，即，·······················································7分由柯西不等式得··························································8分··························································9分，所以（當且僅當，，時等號）．··································10分
 

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