四川省泸州市2021届高三上学期第一次教学质量诊断性考试 文科数学 (含答案)

泸州市高2018级第一次教学质量诊断性考试数  学（文科）   本试卷分第I卷（选择题）和第II卷（非选择题）两部分. 第I卷1至2页，第II卷3至4页.共150分.考试时间120分钟. 注意事项：1. 答题前，先将自己的姓名、准考证号填写在试卷和答题卡上，并将准考证号条形码粘贴在答题卡上的指定位置。2. 选择题的作答：每小题选出答案后，用2B铅笔把答题卡上对应题的答案标号涂黑.3. 填空题和解答题的作答：用签字笔直接答在答题卡上对应的答题区域内，作图题可先用铅笔绘出，确认后再用0.5毫米黑色签字笔描清楚，写在试题卷、草稿纸和答题卡上的非答题区域均无效。4.考试结束后，请将本试题卷和答题卡一并上交。第I卷 （选择题  共60分）一、 选择题：本大题共有12个小题，每小题5分，共60分.每小题给出的四个选项中，只有一项是符合要求的．1．已知集合，，则A． B． C． D．2．“”是“”的 A．充分不必要条件                   B．必要不充分条件C．充要条件                     D．既不充分也不必要条件3．已知，，，则a，b，c的大小关系是A． B． C． D．4．我国的5G通信技术领先世界，5G技术的数学原理之一是著名的香农(Shannon)公式，香农提出并严格证明了“在被高斯白噪声干扰的信道中，计算最大信息传送速率的公式”，其中是信道带宽（赫兹），是信道内所传信号的平均功率（瓦），是信道内部的高斯噪声功率（瓦），其中叫做信噪比．根据此公式，在不改变的前提下，将信噪比从提升至，使得大约增加了，则的值大约为(参考数据:)   A．1559 B．3943 C．1579 D．2512 5．下列函数中，分别在定义域上单调递增且为奇函数的是A．      B．  C．     D．6．右图为某旋转体的三视图，则该几何体的侧面积为A．      B．         C．       D．7．已知两点，是函数与轴的两个交点，且两点A，B间距离的最小值为，则的值为 A．2 B．3 C．4 D．58．函数（其中e是自然对数的底数）的图象大致为 A． B． C． D．9．已知四棱锥中，四边形是边长为2的正方形，且平面，则该四棱锥外接球的表面积为A． B． C． D．10. 定义在R上的函数满足，，当时，，则函数的图象与图象的交点个数为A．1 B．2 C．3 D．4 11．在长方体中，，分别为，的中点，，分别为，的中点，则下列说法错误的是A. 四点B、D、E、F在同一平面内B. 三条直线，，有公共点C. 直线上存在点使，，三点共线D. 直线与直线OF不是异面直线12．已知函数，若存在实数且，使，则实数a的取值范围为 A． B． C． D．第II卷 （非选择题　共90分）注意事项：(1)非选择题的答案必须用0.5毫米黑色签字笔直接答在答题卡上，作图题可先用铅笔绘出，确认后再用0.5毫米黑色签字笔描清楚，答在试题卷和草稿纸上无效.(2)本部分共10个小题，共90分.二、填空题（本大题共4小题，每小题5分，共20分．把答案填在答题纸上）13．已知函数,则的值___________．14．函数的最大值为___________．15．在平面直角坐标系中，角与角均以Ox为始边，它们的终边关于y轴对称．若，则___________．16．已知直四棱柱的所有棱长均为4，且，点E是棱的中点，则过E且与垂直的平面截该四棱柱所得截面的面积为           ．三、解答题：共70分。解答应写出文字说明、证明过程或演算步骤。第17～21题为必考题，每个试题考生都必须作答。第22、23题为选考题，考生根据要求作答。（一）必考题：共60分。17．（本题满分12分）已知函数．（Ⅰ）若，求的值；（Ⅱ）若函数图象上所有点的纵坐标保持不变，横坐标变为原来的倍得到函数的图象，求函数在上的值域．18．（本题满分12分）已知曲线在点处的切线方程为．（Ⅰ）求，b的值；（Ⅱ）判断函数在区间上零点的个数，并证明． 19.（本题满分12分）在中，角，，的对边分别为，，，已知．（Ⅰ）求A；（Ⅱ）已知，，边BC上有一点D满足，求．20.（本题满分12分）如图，在四棱锥S—ABCD中，底面ABCD是菱形，是线段上一点(不含），在平面内过点作//平面交于点．（Ⅰ）写出作点P、GP的步骤(不要求证明)；（Ⅱ）若，，P是SD的中点，求三棱锥的体积．21．（本题满分12分）已知函数，其中，是自然对数的底数.（Ⅰ）当时，求函数在上的最小值；（Ⅱ）设关于x的不等式对恒成立时的最大值为（，），求的取值范围.（二）选考题：共10分。请考生在第22、23题中任选一题作答，如果多做，则按所做的第一题计分。22．（本题满分10分）选修4-4：坐标系与参数方程在平面直角坐标系中，曲线是圆心在（0，2），半径为2的圆，曲线的参数方程为（为参数且），以坐标原点为极点，轴正半轴为极轴建立极坐标系． (Ⅰ) 求曲线的极坐标方程；（Ⅱ）若曲线与两坐标轴分别交于两点，点为线段上任意一点，直线与曲线交于点（异于原点），求的最大值．23．（本题满分10分）选修4-5：不等式选讲若且，已知有最小值为. (Ⅰ) 求的值；（Ⅱ）若使不等式成立，求实数的取值范围. 泸州市高2018级第一次教学质量诊断性考试数  学（文科）参考答案及评分意见评分说明：1．本解答给出了一种或几种解法供参考，如果考生的解法与本解答不同，可根据试题的主要考查内容比照评分参考制订相应的评分细则．2．对计算题，当考生的解答在某一步出现错误时，如果后继部分的解答未改变该题的内容和难度．可视影响的程度决定后继部分的给分，但不得超过该部分正确解答应得分数的一半；如果后继部分的解答有较严重的错误，就不再给分．3．解答右侧所注分数，表示考生正确做到这一步应得的累加分数．4．只给整数分数，选择题和填空题不给中间分．一、选择题题号123456789101112答案BAAC DABA CC DD二、填空题：13．3；   14．0；    15.；    16．.三、解答题：17．解：（Ⅰ）因为  ····························································1分，··························································2分因为，所以，··················································3分所以，·······················································4分即，·························································5分所以；·······················································6分（Ⅱ）图象上所有点横坐标变为原来的倍得到函数的图象，所以函数的解析式为，···········································8分 因为，所以， ·················································9分 所以，······················································11分故在上的值域为．···············································12分18．解：（Ⅰ）因为，·······················································2分所以，·························································3分又因为，·······················································4分点处的切线方程为．所以,··························································5分；····························································6分（Ⅱ）在上有且只有一个零点，··········································7分因为，························································8分当时，，·······················································9分所以在上为单调递增函数且图象连续不断，·····························10分因为，，······················································11分所以在上有且只有一个零点．·······································12分19．解：（Ⅰ）因为，由正弦定理得，··················································2分因为，所以，···················································3分所以，························································4分因为，所以，所以，························································5分所以，所以．···················································6分（Ⅱ）解法一：设的边上的高为，的边上的高为，因为，························································7分所以，························································8分所以，是角的内角平分线，所以，·····································9分因为，可知，···················································10分所以，························································11分所以.·························································12分解法二：设，则，···························································7分因为，，所以，································8分所以，································9分所以，，因为，所以，···················································10分，可知，······················································11分所以，所以.·························································12分解法三：设，，则，在中，由及余弦定理可得：，所以，························································7分因为，可知，···················································8分在中，即，··························································9分在中，，······················································10分即，·························································11分所以．························································12分20.解：（Ⅰ）第一步：在平面ABCD内作GH‖BC交CD于点H；··························2分 第二步：在平面SCD内作HP‖SC交SD于P；·······························4分第三步：连接GP，点P、GP即为所求．··································5分（Ⅱ）因为是的中点，， 所以是的中点，而，所以是的中点，···················································6分所以，连接，交于，连，设在底面的射影为，因为，所以，································7分即为的外心，所以与重合，···························8分因为，，所以，································9分所以，·······························10分因为//平面，···························11分所以．·························································12分21.解：（Ⅰ）当时，，·······················································1分所以，·························································2分因为,由得，·························································3分所以，或，所以在上单减，上单增，············································4分所以函数在上的最小值为；··········································5分（Ⅱ）原不等式.······················································6分因,,所以,令,····························································7分即，令，即，所以在上递增；··················································8分①当即时,因为,所以， 当,,即，所以在上递增，所以，故，························································9分②当即时，因为,,即，所以在上递减，所以，故·························································10分③当即时,又在上递增，所以存在唯一实数,使得,即，则当时，即，当时即，故在上减，上增，所以.························································11分所以，设（），则，所以在上递增,所以.综上所述.·····················································12分22.解： (Ⅰ) 解法一：设曲线与过极点且垂直于极轴的直线相交于异于极点的点E，且曲线上任意点F，边接OF，EF，则OF⊥EF，              2分在△OEF中，，··················································4分解法二：曲线的直角坐标方程为，·····································2分即， 所以曲线的极坐标方程为；······································4分（Ⅱ）因曲线的参数方程为与两坐标轴相交，所以点，·······················································6分所以线段极坐标方程为，···········································7分，,······························8分，····························································9分当时取得最大值为．··············································10分23.解：(Ⅰ)由······························································2分，          解得或（舍去），······················································4分当且仅当时取得“=，          即的最小值为．························································5分（Ⅱ）由，，··························································7分因使不等式成立，所以即，····························································9分即的取值范围是··················································10分