2021福建省漳州市高三毕业班数学适应性测试题答案

这是一份2021福建省漳州市高三毕业班数学适应性测试题答案，共12页。试卷主要包含了C 2，A 8，因为，，所以，等内容，欢迎下载使用。
2021届福建省漳州市高三毕业班数学适应性测试题答案（考试时间：120分钟；满分：150分）本试卷分第Ⅰ卷和第Ⅱ卷两部分。第Ⅰ卷为选择题，第Ⅱ卷为非选择题。第Ⅰ卷（选择题60分）一．单项选择题（本大题共8小题，每小题5分，共40分.在每小题给出的四个选项中，只有一项是符合题目要求的.）1．C        2．D        3．D         4．A        5． B       6．C7．A        8．C 【解析】1．集合，，不满足，则A错；，则B错；，则C正确；，则D错.故选C.2．则，复数在复平面上对应的点为，故复数在复平面上对应的点位于第四象限，故选D.3．依题意可知在角的终边上，所以，故选D．4．点是抛物线内的一点，设点在抛物线准线上的射影为，根据抛物线的定义可知  ，要求的最小值，即求的最小值. 当，，三点共线时，  取到最小值. 故选A.5．由题意得，恰好有6段圆弧或有段圆弧与直线相交时，才恰有个交点，每段圆弧的圆心角都为，且从第1段圆弧到第段圆弧的半径长构成等差数列：，，，当得到的“螺旋蚊香”与直线恰有个交点时，“螺旋蚊香”的总长度的最大值为．故选B.   6． 由题意，以所在直线为x轴，的垂直平分线为y轴建立坐标系，由于，，则，，故又点是的重心，则，，，故选C7．，即为函数与的图象交点的横坐标，即为函数与的图象交点的横坐标，即为函数与的图象交点的横坐标在同一坐标系中画出图象，可得.故选A.8．如图，点，，，，，分别是边，，，，，的中点，这两个正四面体公共部分为多面体．三棱锥是正四面体，其棱长为正四面体棱长的一半，则这两个正四面体公共部分的体积为.   故选C.二、多项选择题（本大题共4小题，每小题5分，共20分。在每小题给出的选项中，有多项符合题目要求。全部选对的得5分，有选错的得0分，部分选对的得3分。）9．ACD              10．BCD           11．ABD          12．BD   【解析】9． 由于小王选择的是每月还款数额相同的还贷方式，故可知2020年用于房贷方面的支出费用跟2017年相同，D错；设一年房贷支出费用为，则可知2017年小王的家庭收入为，2020年小王的家庭收入为，，小王一家2020年的家庭收入比2017年增加了50% ，C错；2017、2020年用于饮食的支出费用分别为，A错；2017、2020年用于其他方面的支出费用分别是，B对．故选ACD．10．由已知，得. 若，则，不满足，故A错；    由，故B正确；当时，且，则，，    所以，故C正确；当时，且，则，，所以    ，所以，则，故D正确. 故选BCD.11．，A正确；已知所以即，D正确；若为锐角三角形，所以 ，若为直角三角形或钝角三角形时可类似证明，B正确；，所以，C错．故选ABD.12．因为的定义域为，，所以是奇函数，但是，所以不是周期为的函数，故A错误；当时，，，单调递增，当时，，，单调递增，且在连续，故在单调递增，故B正确；当时，，，令得，，当时，，令得，,因此，在内有20个极值点，故C错误；当时，，设，所以，令， ，，单调递增，，所以，在单调递增.当趋近于时，趋近于， 所以，故D正确. 故选BD.三．填空题（本大题共4小题，每小题5分，共20分.）13． 14．          15．  16．，【解析】13．展开式的通项为令，则， 所以的展开式中，的系数为故答案为.14．由已知，当时，，不等式等价于又定义在R上的偶函数，所以，所以或，解得或则不等式的解集为故答案为.15．因为，，所以，又因为是等腰直角三角形，，，所以，因为，又，所以，所以．又，所以根据题意可知异面直线与所成角为，根据余弦定理得，故答案为.     16．如图所示建立平面直角坐标系，设的中点为，则由双曲线的对称性知，，所以，所以，可得，，；   的焦距为，所以，. 设，则，又由，得，所以，在中，由余弦定理得，，即，解得，即．故答案为，.四、解答题（本大题共6小题，共70分。解答应写出文字说明、证明过程或演算步骤.）17．解：（1）因为，，所以，因为，所以，······················································2分又因为，所以．·····················································3分所以的面积．······················································5分（2）由（1）可得，所以                 ·························································8分因为，所以，······················································9分所以当时，有最大值1．··············································10分18．解：（1）设等比数列的公比为，根据已知条件， ，，依次构成等差数列，所以，则，························································2分因为，所以，解得···················································4分由，即，所以，解得·················································5分所以····························································6分（2）选①·························································7分································································9分 ·······························································12分选②·····························································7分（*）································································9分可得·······························································11分所以···························································12分选③ ····························································9分 ·······························································12分19．解：（1）取的中点为，连接，则////，且·························································2分所以四边形为平行四边形， 所以//，又面，面所以//面.··························································4分（2）面，则，，是圆柱底面的直径，是弧的中点，所以，为中点，则以为原点，，，分别为轴，轴，轴的正方向建立空间直角坐标系，如图．···························································5分设，则，，，·····················································6分，设平面的一个法向量为，则，，取，则，，则······························································8分，设平面的一个法向量为，则，，解得，取，则， 则·····························································10分所以.···························································11分所以锐二面角的余弦值为.···········································12分          20．解：（1）································································2分································································3分（2）由题意样本方差，故，所以，·····································4分由题意，该厂生产的产品为正品的概率.所以//面.·························································6分（3）所有可能取值为，，，.··········································7分                        ···························································9分随机变量的分布列为0123·······························································11分·······························································12分21．解：（1），，，令，得当，，在单调递减当，，在单调递增所以是的极小值点同时也是最小值点，即································2分当，即时，在上没有零点；当，即时，在上只有1个零点；········································3分因为，所以只有一个零点，又因为，取，，得当，，在单调递增当，，在单调递减，所以对，，所以，即所以，所以内只有一个零点，所以在上有两个零点.···············································5分综上所述，当时，在上有两个零点；当时，函数在上没有零点；当时，函数在上有一个零点.··········································6分（2）方法一：恒成立，即································································7分所以构造，所以，在上单调递增只需，即恒成立····················································8分令， ····························································9分当时，，所以在单调递减；当时，，所以在单调递增，所以，即·························································11分又，所以. ·······················································12分方法二：，有，则当时，····················································7分令，所以在单调递减，··············································8分注意到，所以.（必要性） ··············································9分下面证明（）令，当，，所以在上单调递增当，，所以在上单调递减所以，    即对，，即（）得证.因为，所以，即，即.当时，··························································11分.（充分性）························································12分方法三：，即恒成立································································7分即，恒成立，·····················································8分注意到在单调递增, ················································9分当时，，所以···························································10分当时，注意到，存在，使得，矛盾··········································11分综上，.·························································12分22．解：（1）由题意可得：，解得 ，，·································2分则椭圆方程为·····················································3分（2）设直线的方程为，设，由，整理得  ·······························································4分椭圆的左、右顶点分别为，，直线方程为：，又直线与直线交于点，则， ···········································5分因为,都存在，所以要证，，三点共线，只需证·····························6分只需证只需证只需证只需证而 故，，三点共线.···················································8分（3）由（2）可得·······························································10分令（），则令，函数在区间单调递增，即当，即时，取到最大值.···········································12分  欢迎访问“高中试卷网”——http://sj.fjjy.org