2022南昌高三上學(xué)期摸底考試數(shù)學(xué)（理）試題含答案

這是一份2022南昌高三上學(xué)期摸底考試數(shù)學(xué)（理）試題含答案，共11頁(yè)。試卷主要包含了考生必須保證答題卡整潔，直線，，則“”是“”的，已知向量，，則，已知，且，則的值為，函數(shù)的圖像大致為，已知數(shù)列滿足，則的前20項(xiàng)和等內(nèi)容，歡迎下載使用。
南昌市2022屆高三摸底測(cè)試卷理科數(shù)學(xué)本試卷共4頁(yè)，23小題，滿分150分0考試時(shí)間120分鐘．注意事項(xiàng)：1．答卷前，考生務(wù)必將自己的姓名、準(zhǔn)考證號(hào)填涂在答題卡上，并在相應(yīng)位置貼好條形碼．2．作答選擇題時(shí)，選出每小題答案后，用2B鉛筆把答題卡上對(duì)應(yīng)題目的答案信息涂黑、如需改動(dòng)，用橡皮擦干凈后，再選涂其它答案．3．非選擇題必須用黑色水筆作答，答案必須寫(xiě)在答題卡各題目指定區(qū)域內(nèi)相應(yīng)位置上；如需改動(dòng)，先劃掉原來(lái)答案，然后再寫(xiě)上新答案，不準(zhǔn)使用鉛筆和涂改液．不按以上要求作答無(wú)效．4．考生必須保證答題卡整潔．考試結(jié)束后，將試卷和答題卡一并交回．一．選擇題：本題共12小題，每小題5分，共60分．在每小題給出的四個(gè)選項(xiàng)中，只有一項(xiàng)是符合題目要求的．1．集合的元素個(gè)數(shù)為（    ）A．3 B．4 C．5 D．62．若z為純虛數(shù)，且，則（    ）A．i B．-i C．2i D．-2i3．為數(shù)列的前n項(xiàng)和，若，，則（    ）A． B． C．10 D．4．設(shè)F為拋物線焦點(diǎn)，直線，點(diǎn)A為C上任意一點(diǎn)，過(guò)點(diǎn)A作于P，則（    ）A．3 B．4 C．2． D．不能確定5．直線，，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6．已知向量，，則（    ）A． B． C． D．57．某市出臺(tái)兩套出租車計(jì)價(jià)方案，方案一：2公里以內(nèi)收費(fèi)8元（起步價(jià)），超過(guò)2公里的部分每公里收費(fèi)3元，不足1公里按1公里計(jì)算；方案二：3公里以內(nèi)收費(fèi)12元（起步價(jià)），超過(guò)3公里不超過(guò)10公里的部分每公里收費(fèi)2.5元，超過(guò)10公里的部分每公里收費(fèi)3.5元，不足1公里按1公里計(jì)算．以下說(shuō)法正確的是（    ）A．方案二比方案一更優(yōu)惠B．乘客甲打車行駛4公里，他應(yīng)該選擇方案二C．乘客乙打車行駛12公里，他應(yīng)該選擇方案二D．乘客丙打車行駛16公里，他應(yīng)該選擇方案二8．已知，且，則的值為（    ）A． B． C． D．9．函數(shù)的圖像大致為（    ）A．B．C．D．10．已知數(shù)列滿足，則的前20項(xiàng)和（    ）A． B． C． D．11．已知雙曲線的左、右焦點(diǎn)分別為、，過(guò)的直線l與C的左、右支分別相交于M、N兩點(diǎn)，若，，則雙曲線的離心率為（    ）A． B． C．2 D．12．已知函數(shù)，若，若點(diǎn)不可能在曲線C上，則曲線C的方程可以是（    ）A． B．C．  D．二．填空題：本題共4小題，每小題5分，共20分．13．某企業(yè)三個(gè)分廠生產(chǎn)同一種電子產(chǎn)品，三個(gè)分廠的產(chǎn)量分布如圖所示．現(xiàn)在用分層抽樣方法從三個(gè)分廠生產(chǎn)的產(chǎn)品中共抽取100件進(jìn)行使用壽命的測(cè)試，測(cè)試結(jié)果為第一、二、三分廠取出的產(chǎn)品的平均使用壽命分別為1020小時(shí)、980小時(shí)、1030小時(shí)，估計(jì)這個(gè)企業(yè)生產(chǎn)的產(chǎn)品的平均使用壽命為________小時(shí)．14．若的展開(kāi)式中共有7項(xiàng)，則常數(shù)項(xiàng)為________（用數(shù)字作答）．15．執(zhí)行如下框圖，若輸出的，則輸入x的取值范圍為________．16．正四棱錐，底面四邊形為邊長(zhǎng)為2的正方形，，其內(nèi)切球?yàn)榍?/span>G，平面過(guò)與棱，分別交于點(diǎn)M，N，且與平面所成二面角為30°，則平面截球G所得的圖形的面積為________．三．解答題：共70分．解答應(yīng)寫(xiě)出文字說(shuō)明、證明過(guò)程或演算步驟．第17～21題為必考題，每個(gè)試題考生都必須作答；第22、23題為選考題，考生根據(jù)要求作答．（一）必考題：共60分．17．（12分）在中，角A，B，C所對(duì)的邊分別為a，b，c，，．（Ⅰ）求的值；（Ⅱ）已知的面積為，求b邊．18．（12分）如圖在四棱錐中，底面為正方形，為等邊三角形，E為中點(diǎn)，平面平面．（I）求證：平面；（II）求二面角的余弦值．19．（12分）己知橢圓，，分別為橢圓的左、右焦點(diǎn)，O為坐標(biāo)原點(diǎn)，P為橢圓上任意一點(diǎn)．（Ⅰ）若，求的面積；（Ⅱ）斜率為1的直線與橢圓相交于A，B兩點(diǎn)，，求直線的方程．20．（12分）已知函數(shù)．（Ⅰ）若函數(shù)在定義域上單調(diào)遞增，求實(shí)數(shù)a的取值范圍；（Ⅱ）若函數(shù)存在兩個(gè)極值點(diǎn)，，求實(shí)數(shù)a的取值范圍，并比較與的大?。?/span>21．（12分）甲、乙、丙、丁、戊五位同學(xué)參加一次節(jié)日活動(dòng)，他們都有機(jī)會(huì)抽取獎(jiǎng)券．墻上掛著兩串獎(jiǎng)券袋（如圖），A，B，C，D，E五個(gè)袋子分別裝有價(jià)值100，80，120，200，90（單位：元）的獎(jiǎng)券，抽取方法是這樣的：每個(gè)同學(xué)只能從其中一串的最下端取一個(gè)袋子，得到其中獎(jiǎng)券，直到禮物取完為止．甲先取，然后乙、丙、丁、戊依次取，若兩串都有禮物袋，則每個(gè)人等可能選擇一串?。?/span>（Ⅰ）求丙取得的禮物券為80元的概率；（Ⅱ）記丁取得的禮物券為X元，求X的分布列及其數(shù)學(xué)期望．（二）選考題：共10分．請(qǐng)考生在第22、23題中任選一題作答，如果多做，則按所做的第一題計(jì)分．22．（10分）選修4-4：坐標(biāo)系與參數(shù)方程在平面直角坐標(biāo)系中，直線l的參數(shù)方程為（t為參數(shù)），以坐標(biāo)原點(diǎn)O為極點(diǎn)，x軸正半軸為極軸，建立極坐標(biāo)系，曲線C的極坐標(biāo)方程為．（Ⅰ）求直線l的普通方程和曲線C的直角坐標(biāo)方程；（Ⅱ）設(shè)，直線l與曲線C的交點(diǎn)為M，N，求．23．（10分）選修4-5：不等式選講已知函數(shù)．（Ⅰ）當(dāng)時(shí)，求不等式的解集；（Ⅱ）若對(duì)任意，都有恒成立，求實(shí)數(shù)a的取值范圍． 2022屆高三摸底測(cè)試卷理科數(shù)學(xué)參考答案及評(píng)分標(biāo)準(zhǔn)一、選擇題：本大題共12個(gè)小題，每小題5分，共60分，在每小題給出的四個(gè)選項(xiàng)中，只有一項(xiàng)是符合題目要求的．題號(hào)123456789101112答案CCCAAACCBDBC二．填空題：本大題共4小題，每小題5分，滿分20分．13．1015 14．240 15． 16．三．解答題：共70分．解答應(yīng)寫(xiě)出文字說(shuō)明、證明過(guò)程或演算步驟．第17題-21題為必考題，每個(gè)試題考生都必須作答．第22題、23題為選考題，考生根據(jù)要求作答．17．【解析】（Ⅰ）由正弦定理，（其中R為外接圓的半徑），所以，，，代入已知條件可得：，····································································2分所以，即，·············································································4分，故．·················································································6分（Ⅱ）由已知可得，所以的面積為，故，解得，．···········································································9分所以，即．············································································12分18．【解析】（Ⅰ）連接交于點(diǎn)O，連接、，因?yàn)?/span>為等邊三角形，所以，因?yàn)榈酌?/span>為正方形，所以，因?yàn)?/span>，所以平面，········································································3分又平面，所以，因?yàn)槠矫?/span>平面，平面平面，所以平面，因?yàn)?/span>E為中點(diǎn)，所以，則平面．······························································5分（Ⅱ）如圖，以為x軸，為y軸，為z軸，建立空間直角坐標(biāo)系，設(shè)，則，所以，，，，則，，，因?yàn)槠矫?/span>平面，所以平面的法向量為，····································································7分設(shè)平面的法向量為，則，所以，所以，···········································································9分所以,·················································································11分所以二面角的余弦值為．··································································12分19．【解析】（Ⅰ）由題意，解得，，····························································2分又，所以，即，··················································································4分所以；·················································································5分（Ⅱ）直線斜率為1，設(shè)直線方程，，，由，消元得，得··········································································7分又，知，即·············································································9分而所以，，得，滿足，所以直線的方程或．·····································································12分20．【解析】（Ⅰ）由得······································································2分由題在恒成立，即在恒成立而，所以；·············································································5分（Ⅱ）由題意知，，是方程在內(nèi)的兩個(gè)不同實(shí)數(shù)解，令，注意到，其對(duì)稱軸為直線，故只需，解得，即實(shí)數(shù)a的取值范圍為；···································································8分由，是方程的兩根，得，，因此，···················································································10分又，所以，即得證．··············································································12分21．【解析】（Ⅰ）由題意知，列舉如下：····························································2分所以丙取得的禮物券為80元的概率；（Ⅱ）如下圖，所以X的可能取值為100，80，200，90，又因?yàn)?/span>；；；；所以分布列為：··········································································8分（每個(gè)概率1分）X8090100200P所以．22．【解析】（Ⅰ）直線l的參數(shù)方程為（t為參數(shù)），轉(zhuǎn)換為普通方程為········································································2分曲線C的極坐標(biāo)方程為，根據(jù)轉(zhuǎn)換為直角坐標(biāo)方程為．·······························································5分（Ⅱ）易知直線l的參數(shù)方程標(biāo)準(zhǔn)形式為代入到中，得到；設(shè)M，N所對(duì)應(yīng)的參數(shù)分別為，，則，，·················································································8分所以．················································································10分23．【解析】（Ⅰ）因?yàn)?/span>，所以，當(dāng)時(shí)，，所以；當(dāng)時(shí)，，所以，綜上不等式的解集為．····································································5分（II）因?yàn)?/span>，···········································································8分當(dāng)時(shí)，在單調(diào)遞增；當(dāng)時(shí)，；所以函數(shù)的最小值是a，所以．·····························································10分