高2023級高三第二次教學質(zhì)量診斷性模擬考試數(shù)學試題注意事項：答卷前，考生務必用黑色字跡鋼筆或簽字筆將自己的姓名、考生號、考場號和座位號填寫在答題卡上。2．考生必須保持答題卡的整潔。第=1\*ROMANI卷選擇題（58分）一、選擇題：本題共8小題，每小題5分，共40分。在每小題給出的四個選項中，只有一項是符合題目要求的。1．已知集合，則A． B．C． D．2．下列命題為真命題的是A．若，則 B．若，則C．若，則 D．若，則3．已知，則“”是“”的A．充分不必要條件 B．必要不充分條件C．充要條件 D．既不充分也不必要條件4．記為等差數(shù)列的前n項和.若則A． B． C． D．5．已知，且滿足，則在上的投影向量為A． B． C． D．6．已知的展開式中，前三項的系數(shù)依次成等差數(shù)列，則展開式中二項式系數(shù)最大的項是A． B． C． D．7．某電視臺計劃在春節(jié)期間某段時間連續(xù)播放6個廣告，其中3個不同的商業(yè)廣告和3個不同的公益廣告，要求第一個和最后一個播放的必須是公益廣告，且商業(yè)廣告不能3個連續(xù)播放，則不同的播放方式有A．144種 B．72種 C．36種 D．24種8．已知的定義域為，且，則A． B． C． D．二、選擇題：本題共3小題，每小題6分，共18分。在每小題給出的選項中，有多項符合題目要求。全部選對的得6分，部分選對的得部分分，有選錯的得0分。9．若z是非零復數(shù)，則下列說法正確的是A．若，則 B．若，則C．若，則 D．若，則10．某校在運動會期間進行了一場“不服來戰(zhàn)”對抗賽，由籃球?qū)I(yè)的1名體育生組成甲組，3名非體育生的籃球愛好者組成乙組，兩組進行對抗比賽.具體規(guī)則為甲組的同學連續(xù)投球3次，乙組的同學每人各投球1次.若甲組同學和乙組3名同學的命中率依次分別為，則A．乙組同學恰好命中2次的概率為B．甲組同學恰好命中2次的概率小于乙組同學恰好命中2次的概率C．甲組同學命中次數(shù)的方差為D．乙組同學命中次數(shù)的數(shù)學期望為11．在三棱錐中，平面平面，，則A．三棱錐的體積為1B．點到直線AD的距離為C．二面角的正切值為2D．三棱錐外接球的球心到平面的距離為第=2\*ROMANII卷非選擇題（92分）三、填空題：本題共3小題，每小題5分，共15分．12．計算：．13．函數(shù)（）的最大值是．14．過雙曲線：右焦點作直線，且直線與雙曲線的一條漸近線垂直，垂足為A，直線與另一條漸近線交于點B．且點A，B位于x軸的異側(cè)，O為坐標原點，若的內(nèi)切圓的半徑為，則雙曲線C的離心率為．四、解答題：本題共5小題，共77分．解答應寫出文字說明、證明過程或演算步驟．15．（13分）某市統(tǒng)計了2025年4月的空氣質(zhì)量指數(shù)（AQI），將其分為，，，的4組，畫出頻率分布直方圖如圖所示.若，稱當天空氣質(zhì)量達標；若，稱當天空氣質(zhì)量不達標.(1)求；(2)從4月的30天中任取2天，求至少有1天空氣質(zhì)量達標的概率；(3)若2025年6月的30天中有8天空氣質(zhì)量達標，請完成下面2×2列聯(lián)表，根據(jù)小概率值的獨立性檢驗，能否認為空氣質(zhì)量是否達標與月份有關聯(lián)？月份空氣質(zhì)量合計達標不達標4月6月合計附：，0.10.050.012.7063.8416.63516．（15分）如圖，直三棱柱的體積為4，的面積為．(1)求A到平面的距離；(2)設D為的中點，，平面平面，求二面角的正弦值．17．（15分）如圖，在中，，，，點D在邊BC的延長線上.(1)求的面積；(2)若，，求CE的長.18．（17分）已知橢圓（）的離心率為，其上焦點與拋物線的焦點重合．

(1)求橢圓的方程；(2)若過點的直線交橢圓于點，同時交拋物線于點（如圖1所示，點在橢圓與拋物線第一象限交點上方），試比較線段與長度的大小，并說明理由；(3)若過點的直線交橢圓于點，過點與直線垂直的直線交拋物線于點（如圖2所示），試求四邊形面積的最小值．19．（17分）設函數(shù)，曲線在點處的切線方程為．(1)求的值；(2)設函數(shù)，求的單調(diào)區(qū)間；(3)求的極值點個數(shù)．

高2023級高三第二次教學質(zhì)量診斷性模擬考試數(shù)學試題參考答案一．單選題題號12345678答案CBABDCBB二．多選題題號91011答案BCDBCDACD三．填空題12．13．114．四．解答題15．解：（1）依題意得，，·······································3分解得．··············································································4分（2）由頻率分布直方圖知，4月份的空氣質(zhì)量達標的天數(shù)為：，·································5分則4月份的空氣質(zhì)量不達標的天數(shù)為：，···············································6分則任取2天，至少有1天空氣質(zhì)量達標的概率為：．··································8分（3）列聯(lián)表如下：···········································································10分月份空氣質(zhì)量合計達標不達標4月1218306月82230合計204060零假設：空氣質(zhì)量是否達標與月份無關，則·····································12分所以根據(jù)小概率值的獨立性檢驗，沒有充分理由推斷假設不成立，故不能認為空氣質(zhì)量是否達標與月份有關聯(lián)．····················································13分16．解：（1）在直三棱柱中，設點A到平面的距離為h，則，································3分解得，·················································································5分所以點A到平面的距離為；····························································6分（2）取的中點E,連接AE,如圖，因為，所以,又平面平面，平面平面，·······································7分且平面，所以平面，在直三棱柱中，平面，由平面，平面可得，，·····································8分又平面且相交，所以平面，·········································9分所以兩兩垂直，以B為原點，建立空間直角坐標系，如圖，·····························10分由（1）得，所以，，所以，則,所以的中點，則，,·····················································11分設平面的一個法向量，則，可取，···········································································12分設平面的一個法向量，則，可取，···········································································13分則，···························································14分所以二面角的正弦值為.···············································15分17．解：（1）中，，··························2分因為，，所以，····················································3分所以，因為，·································································5分所以；···································7分（2）方法1：因為,所以,所以，則.················································································9分方法2：在中，由余弦定理得，因為為線段上靠近的三等分點，所以，·············································································11分因為，所以，···········································································12分因為為銳角，所以，························································13分在中，由余弦定理得，，·····························14分所以.············································································15分18．解：（1）由題意得，即：，又，所以，·······························2分由，得，所以橢圓的方程為.···········································3分（2）由題意得過點的直線的斜率存在，設直線方程為，設，，，，聯(lián)立，消去得：，·············································4分則，，所以.·······································5分拋物線的方程為：，聯(lián)立，消去得：，則，所以，······················································7分所以，·············································8分即.················································································9分（3）設，，，，當直線的斜率存在且不為零時，設直線方程為，則直線方程為，由（2）的過程可知：，·······················································11分由，以替換，可得，·········································12分所以，·································································14分因為，所以，，；················15分當直線的斜率不存在時，，，所以；·····················································16分綜上所述：，所以四邊形面積的最小值為.··································17分19．解：（1）因為，所以，··························1分因為在處的切線方程為，·所以，，·······························································2分則，解得，···························································3分所以.（2）由（1）得，則，················4分令，解得，不妨設，，則，易知恒成立，所以令，解得或；令，解得或；所以在，上單調(diào)遞減，在，上單調(diào)遞增，即的單調(diào)遞減區(qū)間為和，單調(diào)遞增區(qū)間為和.··········7分（3）由（1）得，，由（2）知在，上單調(diào)遞減，在，上單調(diào)遞增，當時，，，即所以在上存在唯一零點，不妨設為，則，此時，當時，，則單調(diào)遞減；當時，，則單調(diào)遞增；所以在上有一個極小值點；·························································10分當時，在上單調(diào)遞減，則，故，所以在上存在唯一零點，不妨設為，則，此時，當時，，則單調(diào)遞增；當時，，則單調(diào)遞減；所以在上有一個極大值點；··························································12分當時，在上單調(diào)遞增，則，故，所以在上存在唯一零點，不妨設為，則，此時，當時，