福建省福州市2015届高中毕业班第二次质量(3月)检测数学(理)试题 Word版含答案

-1-福建省福州市2015届高中毕业班第二次质量检测数学（理）试题（完卷时间：120分钟；满分：150分）注意事项：1．本科考试分试题卷和答题卷，考生须在答题卷上作答，答题前，请在答题卷的密封线内填写学校、班级、准考证号、姓名；2．本试卷分为第1卷（选择题）和第Ⅱ卷（非选择题）两部分，全卷满分150分，考试时间120分钟，第I卷（选择题共50分）一、选择题（本大题共10小题，每小题5分，共50分．在每小题所给的四个选项中有且只有一个选项是正确的．把正确选项涂在答题卡的相应位置上．）1．已知全集等于A．[-2,0）B．[-2,0]C．[0,2）D．（0,2）2．在平面直角坐标系中，已知角的顶点与点O重合，始边与x轴的非负半轴重合，终边一点M的坐标为的值是A．-0.5B。0C．0．5D．13．在等差数列，则a5的值是4．若的大小关系为A．abcB．bacC．bcaD．cba5．执行如图所示的程序框图，输出S的值为A．-lB．1-2-C．0D．-20146．一征棱长为3的正方体内任取一点P，则点P到该正方体的六个面的距离的最小值不大于1的概率为7．“直线l垂直于平面”的一个必要不充分条件是A．直线l与平面内的任意一条直线垂直B．过直线l的任意一个平面与平面垂直C．存在平行于直线l的直线与平面垂直D．经过直线l的某一个平面与平面垂直8．的取值范围9．若函数对于函数10．某医务人员说：“包括我在内，我们社区诊所医生和护士共有16名．无论是否把我算在内．下面说法都是对的，在遮些医务人员中：护士多于医生；女医生多于女护士；女护士多于男护。士；至少有一名男医生，”请你推断说话的人的性别与职业是A．男医生B．男护士C．女医生D．女护士第Ⅱ卷（非选择题共l00分）二、填空题（本大题共5小题，每小题4分，共20分．把答案填在答题卡的相应位置上。）11．已知12．展开式的常数项为280，则正数a=.13．已知抛物线的焦点为F，P是的准线上一点，Q是直线PF与的一个交点．若则直线PF的方程为。14．已知一组正数则数据的平均数为。-3-15．已知函数，有下列四个结论：④函数的图象上至少存在三个点，使得该函数在这些点处的切线重合，其中正确结沦的序号是（请把所有正确结论的序号都填上）．三、解答题（本大题共6小题，共80分．解答应写出文字说明、证明过程或演算步骤．）16．（本小题满分13分）已知函数的图象与直线y=2的相邻两个交点之间的距离为（I）求函数的单凋递增区间；（Ⅱ）设△ABC的内角A,B,C所对的边分别是，求角B的大小．17．（本小题满分13分）调查表明，中年人的成就感与收入、学历、职业的满意度的指标有极强的相关性，现将这三项的满意度指标分别记为并对它们进行量化：0表示不满意，1表示基本满意，2表示满意，再用综合指标的值评定中年人的成就感等级：若，则成就感为一级；若2则成就感为二级；若，则成就感为三级，为了了解目前某群体中年人的成就感情况，研究人员随机采访了该群体的10名中年人，得到如下结果：（I）茌这10名被采访者中任取两人，求这两人的职业满意度指标z相同的概率；（Ⅱ）从成就感等级是一级的被采访者中任取一人，．其综合指标为a，从成就感等级不是一级的被采访者中任取一人，其综合指标为b，记随机变量的分布列及其数学期望．18．（本小题满分13分）已知一个空间几何体的直观图和三视图（尺寸如图所示）．-4-（I）设点M为棱肋中点，求证：（Ⅱ）线段加上是否存在一点N，使得直线与平面PCD所成角的正弦值等于手？若存在，试确定点N的位置；若不存在，请说明理由，19．（本小题满分13分）20．（本小题满分14分）已知函数为自然对数的底数．（I）当a=e时，求函数处的切线方程；（Ⅱ）设的大小，并加以证明．21．本题设有（1）（2）（3）三个选考题，每题7分，请考生任选2题作答，满分14分．若多做，则按所做的前两题计分（1）（本小题满分7分）选修4q：矩阵与变换已知矩阵A的逆矩阵（I）求矩阵A；（Ⅱ）求曲线在矩阵A所对应的线性变换作用下所得的曲线方程．-5-（2）（本小题满分7分）选修4-4：坐标系与参数方程己知曲线C1的参数方程为．在平面直角坐标系中，以坐标原点为极点，x轴的非负半轴为极轴建立极坐标系，曲线C2的极坐标方程为（I）把C1的参数方程化为极坐标方程；（Ⅱ）求C1与C2交点的极坐标（3）（本小题满分7分）选修4-5：不等式选讲已知定义在的最小值为3．（I）求a的值；（Ⅱ）求不等式的解集．-6-2015年福州市高中毕业班质量检测理科数学能力测试参考答案及评分细则第Ⅰ卷（选择题共50分）一、选择题（本大题共10小题，每小题5分，共50分．）第Ⅱ卷（非选择题共100分）二、填空题（本大题共5小题，每小题4分，共20分．）11．34i12．213．10xy或10xy14．315．①③④三、解答题（本大题共6小题，共80分．)(Ⅱ)由(Ⅰ)知，π()2sin(2).6fxx在ABC中，因为()2,fA所以π2sin(2)2,6A············································································································7分所以πsin(2)1,6A因为0πA，所以π3A．······························································································9分因为3ab，根据据正弦定理，有sin3sinAB，···················································10分所以πsin3sin3B，所以1sin2B，·············································································11分因为ab，所以AB，所以π03B，······································································12分所以π6B．························································································································13分-7-（Ⅱ）计算10名被采访者的综合指标，可得下表：人员编号1A2A3A4A5A6A7A8A9A10A综合指标4462453513其中成就感是一级的（4w）有：1A、2A、3A、5A、6A、8A，共6名，成就感不是一级的（4w）有4A、7A、9A、10A，共4名．随机变量X的所有可能取值为：1,2,3,4,5．···································································6分113211641(1)4CCPXCC,····································································································7分1111312211647(2)24CCCCPXCC，·················································································8分11111131212111647(3)24CCCCCCPXCC，································································9分(4)PX111111211164CCCCCC18,··············································································10分111111641(5)24CCPXCC，·······························································································11分所以X的分布列为X12345p1472472418124···············································································································································12分所以1771129()123454242482412EX．········································13分-8-18．本小题主要考查空间体的直观图与三视图、直线与平面的平行、线面所成角、探索性问题等基础知识，考查空间想象能力、推理论证能力、运算求解能力，考查数形结合思想、化归与转化思想、函数与方程思想．满分13分．（Ⅰ）证明：（方法一）由三视图知，,,BABPBC两两垂直，故以B为原点，,,BABPBC分别为x轴，y轴，z轴正方向，建立如图所示的空间直角坐标系．…………………1分则1(0,2,0),(2,0,1),(1,1,),(2,1,0),(0,0,1)2PDMEC,所以1=(1,0,)2EM，易知平面ABCD的一个法向量等于(0,1,0)n，………3分所以1=(1,0,)(0,1,0)02EMn,所以EMn，·····················································································································4分又EM平面ABCD，所以EM∥平面ABCD．···································································································5分（方法二）由三视图知，,,BABPBC两两垂直．连结,ACBD，其交点记为O，连结MO,EM．····················································1分因为四边形ABCD为矩形，所以O为BD中点．因为M为PD中点，所以OM∥PB，且12OMPB．………………………2分又因为AE∥PB，且12AEPB，所以AE∥OM,且AE=OM．所以四边形AEMO是平行四边形，所以EM∥AO………………………………………4分因为EM平面ABCD，AO平面ABCD所以EM∥平面ABCD．···································································································5分（Ⅱ）解：当点N与点D重合时，直线BN与平面PCD所成角的正弦值为25．······6分理由如下：因为(2,2,1),(2,0,0)PDCD，设平面PCD的法向量为1111(,,)nxyz，由110,0nPDnCD得1111220,20.xyzx···············································································7分取11y，得平面PCD的一个法向量1(0,1,2)n