广东省部分名校2022-2023学年高一下学期3月大联考数学试题

这是一份广东省部分名校2022-2023学年高一下学期3月大联考数学试题，共10页。试卷主要包含了本试卷主要考试内容，如图，在正六边形中，，已知曲线，下列说法正确的是，已知函数的图象经过点，则等内容，欢迎下载使用。
高一数学注意事项：1．答题前，考生务必将自己的姓名、考生号、考场号、座位号填写在答题卡上。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，再选涂其他答案标号。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，将本试卷和答题卡一并交回。4．本试卷主要考试内容：人教A版必修第一册5.3至第二册6.2。一、选择题：本题共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．（    ）A． B． C． D．2．函数的最小正周期和最大值分别是（    ）A．和3 B．和2 C．和3 D．和23．下列函数为奇函数且在上为减函数的是（    ）A． B． C． D．4．已知函数，则（    ）A．的图象关于点对称 B．的图象关于直线对称C．为奇函数 D．为偶函数5．如图，在正六边形中，（    ）A． B． C． D．6．已知曲线：，：，则下面结论正确的是（    ）A．把上各点的横坐标伸长到原来的2倍，纵坐标不变，再把得到的曲线向右平移个单位长度，得到曲线B．把上各点的横坐标伸长到原来的2倍，纵坐标不变，再把得到的曲线向左平移个单位长度，得到曲线C．把上各点的横坐标缩短到原来的，纵坐标不变，再把得到的曲线向右平移个单位长度，得到曲线D．把上各点的横坐标缩短到原来的，纵坐标不变，再把得到的曲线向左平移个单位长度，得到曲线7．已知两个单位向量和的夹角为120°，则向量在向量上的投影向量为（    ）A． B． C． D．8．为了研究钟表秒针针尖的运动变化规律，建立如图所示的平面直角坐标系，设秒针针尖位置为点．若初始位置为点，秒针从（规定此时）开始沿顺时针方向转动，若点P的纵坐标为y，，则时t的取值范围为（    ）A． B． C． D．二、选择题：本题共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，部分选对的得2分，有选错的得0分．9．下列说法正确的是（    ）A．平行向量不一定是共线向量B．向量的长度与向量的长度相等C．是与非零向量共线的单位向量D．若四边形满足，则四边形是平行四边形10．已知函数的图象经过点，则（    ）A．B．的最小正周期为C．的定义域为D．不等式的解集为，11．已知黄金三角形是一个等腰三角形，其底与腰的长度的比值为黄金比值（即黄金分割值，该值恰好等于），则下列式子的结果等于的是（    ）A． B．C． D．12．如图，在平行四边形中，，，，延长DP交BC于点M，则（    ）A． B．C．  D．三、填空题：本题共4小题，每小题5分，共20分．把答案填在答题卡中的横线上．13．已知单位向量，，满足，则________．14．已知，则________．15．已知函数是定义在上周期为2的奇函数，当时，，则________；当时，________．（本题第一空2分，第二空3分）16．已知函数在上有最大值，无最小值，则的取值范围是________．四、解答题：本题共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，为第二象限角．（1）求的值；（2）求的值。18．（12分）设，是不共线的两个向量，若，，．（1）若，，且，求与的夹角；（2）若A，B，C三点共线，求m的值．19．（12分）已知函数。（1）求的单调递减区间；（2）若，，求的值．20．（12分）某同学用“五点法”画函数在某一个周期内的图象时列表并填入了部分数据，如下表：0x   02 0（1）请将上表数据补充完整，填写在答题卡上相应位置，并写出函数的解析式．（2）将的图象向左平行移动个单位长度，得到的图象。若的图象关于直线对称，求的最小值．21．（12分）在中，．（1）求的值；（2）若，求．22．（12分）在中，，且。（1）求A；（2）已知E为BC的中点，点D为AC上一点，且，BD与AE相交于点P，求． 高一数学参考答案1．A  ．2．D  的最小正周期，最大值为2．3．D  利用函数的图象易知为奇函数且在上为减函数，故选D．4．C  ，，A错误；，B错误；，所以是奇函数，C正确；易知，所以不是偶函数，D错误．5．A  因为六边形为正六边形，所以．6．C  由题可知，把上各点的横坐标缩短到原来的，纵坐标不变，得到函数的图象，再把得到的曲线向右平移个单位长度，得到函数的图象，即曲线．7．D  因为，所以，故向量在向量上的投影向量为．8．B  设y与时间t的函数关系式为，由题意可得，初始位置为，即初相为，故可得，，则，．又函数周期是60（秒）且秒针按顺时针方向旋转，即，所以，即，．令，则，解得．9．BCD  平行向量即共线向量，故A错误．与为相反向量，所以模长相等，故B正确．是与非零向量共线的单位向量，C正确．，所以且，则四边形是平行四边形，D正确．10．BD  由题知，则，因为，所以，A错误．的最小正周期，B正确．令，，则，，所以的定义域为，C错误．令，则，得，，即，，所以不等式的解集为，，D正确．11．BCD  对于A，；对于B，；对于C，；对于D，．故选BCD．12．ACD  因为在平行四边形中，所以，即M为BC的中点，所以，，．13．  由，可得，平方可得，解得．14．  因为，则，所以．15．；    ，当时，则，所以，所以，当时，．16．  ．由题可知，，所以，当时，，所以解得．17．解：因为，为第二象限角，所以，．··················································2分（1）方法一：·················································································4分．················································································5分方法二：·················································································4分．················································································5分（2），，··········································································7分．···············································································10分18．解：（1），····································································1分因为，所以，·······································································3分解得，则，所以与的夹角为．···························································6分（2）因为，，······································································8分且A，B，C三点共线，所以存在，使得，即，··············································10分则，解得．········································································12分19．解：（1）．····································································2分由，，得，，·······································································5分所以的单调递减区间为（）．···························································6分（2）由，得，即．··············································································7分因为，所以，则，，··································································9分故．·············································································12分20．解：（1）根据表中已知数据，得，···················································1分，可得，··········································································2分当时，，解得，·····································································3分所以．············································································4分数据补全如下表：0x0200·················································································6分（2）由（1）知，得．································································8分令，解得，．·······································································9分由于函数的图象关于直线对称，令，解得，．··········································································11分由可知，当时，取得最小值．··························································12分21．解：（1）因为，所以，即．·········································································3分．················································································6分（2）因为，，所以，．·······························································8分故···············································································10分．···············································································12分22．解：（1）根据，可得．····························································2分所以．············································································4分又，所以．·········································································5分（2）因为，，所以，易知．····························································6分因为E为BC的中点，所以，．···························································8分因为，所以，，····································································10分则，所以．···········································································12分