厦门市2017-2018高一下学期质量检测数学试题(终稿)

高一数学试题第1页（共4页）厦门市2017～2018学年度第二学期高一年级质量检测数学试题注意事项：1．答卷前，考生务必将自己的姓名、准考证号填写在答题卡上．2．回答选择题时，选出每小题答案后，用铅笔把答题卡上对应题目的答案标号涂黑．如需改动，用橡皮擦干净后，再选涂其它答案标号．回答非选择题时，将答案写在答题卡上．写在本试卷上无效．3．考试结束后，将答题卡交回．一、选择题：本大题共12小题，每小题5分，共60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．已知为第四象限角，22sin3，则cos等于A．13B．19C．19D．132．已知直线a，b，c，平面，则下列结论错误．．的是A.若//ab，//bc，则//acB.若ab，bc，则//acC.若a，b，则//abD.若//ab，a，则b3．已知扇形的圆心角为56，半径为4，则该扇形的面积为A．53B．103C．203D．4034．已知=(3,1)ax，(,2)bx，若a与b的方向相反，则实数x的值为A．2-B．3C．2-或3D．2或35．已知点(3,2)A和(1,4)B到直线30mxy的距离相等，则实数m的值为A．0或21B．6或21C．6或21D．0或216．已知点(1,2)A，(3,1)B-，则AB在y轴正方向上的投影为A．3B．2C．2-D．3-7．如图，弹簧挂着的小球作上下运动，它在t秒时相对于平衡位置的高度h厘米由关系式2sin()3ht确定．下列结论正确的是A．小球的最高点和最低点相距2厘米B．小球在0t时的高度1hC．每秒钟小球往复运动的次数为2D．从1t到3t，弹簧长度逐渐变长8．榫卯是我国古代工匠极为精巧的发明，是中国古建筑、家具及其它器械的主要结构方式，其特点是在物件上不使用钉子，利用榫卯加固物件．图1所示的榫卯结构由两部分组成，其中一部分结构的三视图如图2所示，网格纸上小正方形的边长为1，则该部分的表面积为高一数学试题第2页（共4页）A．185164B．185194C．105162D．1051929．若圆224260xyxym与y轴的交点,AB位于原点同侧，则实数m的取值范围是A．1mB．5mC．65mD．61m10．如图，正三棱柱111CBAABC的底面边长为3，侧棱长为1，M为AB的中点，则直线1AM与1BC所成角的正弦值是A.74B.34C.32D.7311．如图，在ABC△中，2ABC，AD平分BAC，过点B作AD的垂线，分别交AD，AC于E，F．若6AF，8BC，则=AEA．13210ABACB．1123ABACC．23510ABACD．2153ABAC12．已知()sin()(0,0)fxAxA的图象与直线(0)ymm的三个相邻交点的横坐标分别是21014,,333．当[,]xmA时，)(xf的值域为22[,]33，则A的值是A.23B.43C.2D.4二、填空题：本大题共4小题，每小题5分，共20分.13．过圆22(4)(2)25xy上的点)21(，M作切线l，则l的方程是________．14．已知sin()2cos，则2sin2cos________．15．已知a，b，c均是单位向量，若3abc，则a与b的夹角为________．16．正方体1111DCBAABCD的棱长为1，过1AC的平面截此正方体所得四边形周长的最小值是________．三、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．（10分）已知ABC△中，点(4,3)A，(2,1)B，点C在直线l：220xy上．（1）若C为l与x轴的交点，求ABC△的面积；（2）若ABC△是以AB为底边的等腰三角形，求点C的坐标．高一数学试题第3页（共4页）18．（12分）如图，直四棱柱1111DCBAABCD中，底面ABCD是菱形，60BAD，E是1CC的中点，122AAAB.（1）证明：1AEBD；（2）求直线EA1与平面ABCD所成角的大小．19．（12分）如图，角,的顶点为坐标原点，始边与x轴的非负半轴重合，锐角的终边与单位圆交于点A.（1）若点A的坐标为43(,)55，OA绕原点逆时针旋转4，与角的终边重合，求sin；（2）已知点(0,3)C，(1,0)D，角终边的反向延长线与单位圆交于点B，求当角取何值时，四边形ABCD的面积最大？并求出这个最大面积．20．（12分）一木块如图所示，点G是SAC△的重心，过点G将木块锯开，使截面平行于侧面SBC．（1）在木块上画出符合要求的线，并说明理由；（2）若底面ABC为等边三角形，232SASBSCAB，求截面与平面SBC之间的几何体的体积．高一数学试题第4页（共4页）21．（12分）已知函数()sin()(0,)2fxx的周期为，其图象关于直线4x对称．（1）求()fx的解析式，并画出其在区间0,上的图象；（2）将()fx图象上的点的横坐标缩短至原来的12倍（纵坐标不变），再把所得图象上所有的点向左平移8个单位得到()gx的图象．当0,10x时，求函数()()(21)()1Fxgxafxa(01)a的零点个数．22．（12分）如图，圆C与x轴交于点(1,0)A，(1,0)B，其在x轴下方的部分和半圆221xy(0)y组成曲线．过点A的直线l与的其它两个交点为E，F，且点E在x轴上方．当E在y轴上时，2FAAE．（1）求C的方程；（2）延长EB交于点G．求证：CFG△的面积为定值．高一数学试题第5页（共4页）厦门市2017～2018学年度第二学期高一年级质量检测数学参考答案一、选择题1~5ABCAB；6~10DDCCB；11~12AB第12题参考解答：21014()()()0333fffm，142433T，22T，又10214103333，当1023322x时，fx取最小值，则32222k，2,2kk，sin()cos222fxAxAx2()32Amf．依题意23A，若23A，222ATAm，与23A矛盾，舍去；若23A，则fx在,mA上单调，2TAm，即4A．22033fmfA，()02mAf，则222mAk，4833Ak,kZ，43A．二、填空题13．3450xy14．115．6016．25三、解答题17．本题考查直线平行与垂直的性质、点到直线的距离、两点距离公式以及三角形面积等基础知识，考查运算求解能力，考查方程思想与数形结合的数学思想．本题满分10分．解：（1）法一：点C在直线l上，当0y时，2x，(2,0)C··································································1分2ABk，直线AB的方程为250xy，·····································2分C到直线AB的距离95d············································································3分25AB，·····························································································4分11925=9225ABCSABd···························································5分法二：点C在直线l上，当0y时，2x，(2,0)C··································································1分2ABk，直线AB的方程为250xy，········································2分令0y，则52x，······················································································3分高一数学试题第6页（共4页）15(2)(13)922ABCS······································································5分（2）AB中点的坐标为(3,1)，2ABk······························································6分AB的中垂线方程为11(3)2yx，即250xy····································7分联立250220xyxy，····················································································8分得3274xy，点37(,)24C··············································································10分18．本题考查线面的位置关系、线面角等基础知识，考查逻辑推理、运算求解等能力，考查化归转化的数学思想．本题满分12分．解：（1）证明：连结AC，四边形ABCD是菱形ACBD·····························································1分1AA平面ABCD1AABD·····························································2分又1AAACA························································································3分BD平面1ACEA······················································································5分1AEBD·································································································6分（2）法一：取1AA中点F，连结CF，则1//AECF············································7分CF与平面ABCD所成角等于1AE与平面ABCD所成角又1AA平面ABCDFCA为CF与平面ABCD所成角··························9分又60BAD，1AB3AC····························································10分13tan33AFFCAAC·········································································11分30FCA即1AE与平面ABCD所成角为30····