广东省佛山市第一中学2022-2023学年高一数学下学期第一次教学质量检测试题（Word版附答案）

佛山一中2022~2023学年下学期高一第一次教学质量检测

数学科试题

一、选择题：本大题共8个小题，每小题5分，共40分，在每小题的四个选项中，只有一项是符合题目要求的．

1．    (　　)
A． B． C． D．

2．角的终边经过点，则的值为  (　　)
A． B． C． D．

3．若，则   (　　)
A． B． C．  D．

4．设D为△ABC所在平面内一点，，则  (　　)

A． B．
C． D．

5．已知，则  (　　)

A． B． C． D．

6．如图，若,则的值为                                (　　)

A． B． C． D．

7．如图，扇形OPQ的半径为1，圆心角为，C是扇形弧上的动点，ABCD是扇形的内接矩形，记，矩形ABCD的面积最大值为                            (　　)

A． B． C． D．

8．函数的图象如图所示，为了得到的图象，可以将的图象                 (　　)

A．向右平移个单位长度 B．向右平移个单位长度
C．向左平移个单位长度 D．向左平移个单位长度

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．全部选对得5分，有选错的得0分，部分选对的得2分．

9．下列选项中，与的值相等的是   (　　)

A．  B．
C．  D．

10．设，则   (　　)

A．的最小正周期为 B．是的一条对称轴
C．在上单调递增 D．向右平移个单位后为一个偶函数

11．定义两个非零平面向量的一种新运算：，其中表示的夹角，则对于两个非零平面向量，下列结论一定成立的是                            (　　)

A．  B．
C． D．若，则与平行

12．已知函数，下列关于该函数结论正确的是 (　　)

A．的图象关于直线对称 B．的一个周期是
C．的最大值为2 D．是区间上的增函数

三、填空题：本大题共4小题，每小题5分，共20分．

13．不等式的解集是________．

14．函数的最小值为________．

15．已知，，则________．

16．如图，点C在半径为1，圆心角的扇形OAB的弧上运动．已知，则当时，________；的最大值为________．

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，且．

(1)求的值．

(2)若，，求的值．

18．(12分) 已知，，．求：

(1)与的夹角．

(2)．

19．(12分) 已知函数．

(1)求的最小正周期和对称中心．

(2)求在区间上的最大值和最小值．

20．(12分) 摩天轮是一种大型转轮状的机械建筑设施，游客坐在摩天轮的座舱里慢慢的往上转，可以从高处俯瞰四周的景色(如图1)某摩天轮的最高点距离地面的高度为90米，最低点距离地面10米，摩天轮上均匀设置了36个座舱(如图2)．开启后摩天轮按逆时针方向匀速转动，游客在座舱离地面最近时的位置进入座舱，摩天轮转完一周后在相同的位置离开座舱.摩天轮转一周需要30分钟，当游客甲坐上摩天轮的座舱开始计时．
                

(1)经过t分钟后游客甲距离地面的高度为H米，求的解析式．

(2)问：游客甲坐上摩天轮后多长时间，距离地面的高度恰好为30米？

21．(12分) 己知函数在区间单调，其中，，且．

(1)求图象的一条对称轴．

(2)若，求．

22．(12分) 函数()的部分图象如右图，把函数的图象向右平移个单位，得到函数的图象．

(1)当时，求函数的单调递减区间．

(2)对于，是否总存在唯一的实数，使得成立？若存在，求出实数m的值或取值范围；若不存在，说明理由．

佛山一中2022~2023学年下学期高一第一次教学质量检测

数学参考答案与评分标准

第一部分　选择题答案

第二部分　填空题答案

13．．

14．．

15．．

16．；．(第一空2分，第二空3分)

第三部分　解答题答案与评分标准

17．解：

(1)，且．

．········································································2分

．····························································4分

(2)，，，．·································································6分

．········································································8分

··········································································10分

18．解：

(1)，
，即，·····································································2分

．········································································4分

又的取值范围为，

．········································································6分

(2)········································································8分
化为．·····································································12分

19．解：

(1)因为
··········································································3分

所以的最小正周期．··························································4分
令，解得，
所以的对称中心为 ．··························································6分

(2)当时，，．·······························································8分

当，即时，取得最小值．······················································10分

当，即时，取得最大值．······················································12分

20．解：

(1)由题意知：．······························································2分

，故．·····································································4分
由题可取，
因此，·························································6分

(2)令，则，即．······························································8分
因为，则，所以或，··························································10分
解得或，
故游客甲坐上摩天轮5分钟或25分钟时，距离地面的高度恰好为30米······················12分

21．解：

(1)由题设，的最小正周期．·····················································2分

又因为，
所以为图象的一条对称轴．．···················································4分

(2)由（1）知，故，由，得或．··················································5分

由为的对称轴，
所以．·····································································6分

因为，
所以或．···································································7分
若，则，
即．不存在整数，使得或．·····················································9分
若，则，
即．不存在整数，使得或．·····················································10分
当时，．···································································11分
此时，由，得．·····························································12分

22．解：

(1)由函数图象可知，，，·······················································1分
，．·······································································2分

当时，，．·································································3分
又，故，···································································4分
，，
．········································································5分
由，解得，
可得函数的单调递减区间为·····················································6分

(2)由，得，
由，可得，，
．········································································8分

又，得，所以，
由的唯一性可得：即．························································10分

由，得，解得，
综上所述，当时，使成立．·····················································12分