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    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)

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    这是一份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),共8页。试卷主要包含了选择题,填空题,解答题等内容,欢迎下载使用。

    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的终边经过点,则的值为  (  )
    A B C D

    3,则   (  )
    A B C  D

    4DABC所在平面内一点,,则  (  )

    A B
    C D

    5已知,则  (  )

    A B C D

    6如图,,的值为                                (  )

    A B C D

     

    7如图扇形OPQ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函数图象如图所示,为了得到图象,可以将图象                 (  )

    A向右平移单位长度 B向右平移单位长度
    C向左平移单位长度 D向左平移单位长度

    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,则   (  )

    A最小正周期为 B的一条对称轴
    C上单调递增 D向右平移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,则平行

    12已知函数,下列关于该函数结论正确的是 (  )

    A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半径1,圆心角扇形OAB的弧上运动.已知则当时,________的最大值________

     

    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) 已知,且

    (1)的值

    (2),求的值

    18(12) 已知

    (1)的夹角

    (2)


    19(12) 已知函数

    (1)最小正周期和对称中心

    (2)在区间上的最大值和最小值.

    20(12) 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(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(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) 己知函数在区间单调,其中

    (1)图象的一条对称轴

    (2),求

    22(12) 函数()的部分图象图,把函数图象向右平移单位,得到函数图象

    (1)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

    题号

    1

    2

    3

    4

    5

    6

    7

    8

    9

    10

    11

    12

    答案

    A

    D

    D

    A

    B

    C

    A

    B

    ABC

    AC

    AD

    ABD

     

    部分 填空题答案

    13

    14

    15

    16(第一空2,第二空3)

     

    部分 解答题答案与评分标准

    17解:

    (1),且

    ········································································2

    ····························································4

    (2)·································································6

    ········································································8

    ··········································································10

     

    18解:

    (1)
    ·····································································2

    ········································································4

    取值范围为

    ········································································6

    (2)········································································8
    化为·····································································12

     

    19解:

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

    所以最小正周期··························································4
    ,解得
    所以的对称中心为 ··························································6

    (2)·······························································8

    ,即取得最小值······················································10

    ,即取得······················································12

     

    20解:

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

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

    (2),即······························································8
    因为,则所以··························································10
    解得
    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······················12

     

    21解:

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

    又因为
    所以图象的一条对称轴.···················································4

    (2)由(1)知,故,由,得··················································5

    的对称轴,
    所以·····································································6

    因为
    所以···································································7
    ,则
    .不存在整数,使得·····················································9
    ,则
    .不存在整数,使得·····················································10
    时,···································································11
    此时,由,得·····························································12

     

    22解:

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

    时,·································································3
    ···································································4

    ········································································5
    ,解得
    可得函数的单调递减区间为·····················································6

    (2),得
    ,可得
    ········································································8

    ,得,所以
    的唯一性可得:························································10

    ,得,解得
    综上所述,当时,使成立·····················································12

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