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    江西省赣州市2023届高三数学(理)上学期1月期末考试试题(Word版附解析)

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    这是一份江西省赣州市2023届高三数学(理)上学期1月期末考试试题(Word版附解析),共10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。

    赣州市20222023学年度第一学期期末考试

     高三数学(理科)试卷 20231

    试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分,共150分,考试时间120分钟

    第Ⅰ卷

    一、选择题:本题共12小题,每小题5分,共60.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.集合,则()

    A. B. C. D.

    2.函数()

    A. B.0 C. D.2

    3.若数列是等比数列,且,则()

    A. B. C.62 D.64

    4.为了研究某班学生的右手拃长x(单位:厘米)和身高y(单位:厘米)的关系,从该班随机抽取了12名学生,根据测量数据的散点图可以看出yx之间有线性相关关系,设其回归直线方程为,已知,若某学生的右手拃长为22厘米,据此估计其身高为()

    A.175 B.179 C.183 D.187

    5.若复数a为其共轭复数),定义:.则对任意的复数,有下列命题:

    ,则为纯虚数.

    其中正确的命题个数为()

    A.1 B.2 C.3 D.4

    6.程大位1533~1606,明朝人,珠算发明家.在其杰作《直指算法统宗》里,有这样一道题:荡秋千,平地秋千未起,踏板一尺离地,送行二步与人齐,五尺人高曾记。仕女佳人争,终朝笑语欢,良工高士素好奇,算出索长有几?将其译成现代汉语,其大意是,一架秋千当它静止不动时,踏板离地一尺,将它向前推两步(古人将一步算作五尺)即10尺,秋千的踏板就和人一样高,此人身高5尺,如果这时秋千的绳索拉得很直,请问绳索有多长?()

    A.14 B.14.5 C.15 D.15.5

    7.已知过抛物线C的焦点F的直线lC截得的弦长为8,则坐标原点Ol的距离为()

    A. B. C. D.

    8.展开式的各项系数和为729展开式中的系数为()

    A. B. C.30 D.90

    9.直线与双曲线E交于MN两点,若为直角三角形(其中O为坐标原点),则双曲线E的离心率为()

    A. B. C. D.

    10.已知函数的最小值为a,则实数a的值为()

    A. B. C. D.1

    11.在三棱锥中,,且,则三棱锥的外接球的表面积为()

    A. B. C. D.

    12.已知,则()

    A. B. C. D.

    第Ⅱ卷

    本卷包括必考题和选考题两部分,第1321题为必考题,每个试题考生都必须作答,第2223题为选考题,考生根据要求作答.

    二、填空题:本题共4小题,每小题5分,共20.

    13.已知均为单位向量且夹角为45°,则________.

    14.已知tan,则__________.

    15.如图,是同一平面内的三条平行直线,间的距离是1间的距离是2,等腰直角三角形的三顶点分别在l2上,则的斜边长可以是__________(写出一个即可).

    16.斐波那契,意大利数学家,其中斐波那契数列是其代表作之一,即数列满足,且,则称数列为斐波那契数列.已知数列为斐波那契数列,数列满足,若数列的前12项和为86,则__________.

    三、解答题:共70.解答应写出文字说明、证明过程或演算步骤.1721题为必考题,每个试题考生都必须作答.2223题为选考题,考生根据要求作答.

    17.(本小题满分12分)

    中,角ABC的对边分别为abc,满足.

    1)求B的值;

    2)若边上的高之比为35,且,求的面积.

    18.(本小题满分12分)

    如图,在四棱锥中,,且.

    1)证明:平面平面

    2)求平面与平面夹角的余弦值.

    19.(本小题满分12分)

    设有标号为123nn小球(除标号不同外,其余均一样)和标号为123nn盒子,将这n小球任意地放入这n盒子,每个盒子放一个小球,若i23n号球放入i号盒子,则称该球放对了,否则称放错了.表示放对了的球的个数.

    1)当时,求的概率;

    2)当时,求的分布列与数学期望.

    20.(本小题满分12分)

    已知椭圆C过点且离心率为,过点作两条斜率之和为0的直线CAB两点,CMN两点.

    1)求椭圆C的方程;

    2)是否存在实数使得?若存在,请求出的值,若不存在,说明理由.

    21.(本小题满分12分)

    已知函数其中e为自然对数的底数,且曲线处的切线方程为.

    1)求实数mn的值;

    2)证明:对任意的,有.

    请考生在第2223题中任选一题作答.如果多做,则按所做的第一题计分.

    22.[选修4-4:坐标系与参数方程]

    在平面直角坐标系中,已知直线l的参数方程为t为参数),以坐标原点O为极点,x轴的正半轴为极轴,取相同的单位长度建立极坐标系,曲线C的极坐标方程为.

    1)求直线l的普通方程和曲线C的直角坐标方程;

    2)设点,直线l与曲线C的交点为AB的值.

    23.[选修4-5:不等式选讲]

    已知函数的最小值为m.

    1)求m的值;

    2)设abc为正数,且,求证:.

     


    赣州市20222023学年度第一学期期末考试

    高三理科数学参考答案

    选择题

    题号

    1

    2

    3

    4

    5

    6

    7

    8

    9

    10

    11

    12

    答案

    A

    D

    C

    C

    B

    B

    C

    D

    B

    D

    C

    A

    12.解:由,得,由,即

    ,令

    ,则上单调递增,

    ,即,即,则.

    或由对数平均不等式,即

    .

    填空题

    13.114.15.(三个任选一个);16.8.

    16.解析:斐波那契数列:1123581321345589144…….

    (特征:每三项中前两项为奇数后一项为偶数)

    得:

    同理:

    得:

    ,则.

    解答题

    17.解:(1)由

    由内角和定理得:····································································1

    进而得:···········································································2

    解得:(舍去)····································································4

    从而得.·············································································6

    2由题设知,不妨设······························································7

    由余弦定理得:······································································8

    联立得:···········································································9

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

    ,从而的面积·····································································12

    18.解:1连接,由题设,得.························································1

    ,故,由余弦定理可得·······························································2

    从而有·············································································3

    ,得········································································4

    ,故·············································································5

    ,所以面······································································6

    2)法:由(1)同理可得························································6

    为坐标原点,,分别为xy轴的正方向建立如图所示的空间直角坐标系,

    ········································································7

    1,故面的一个法向量为·······················································8

    设面的法向量为

    结合,令,得········································································9

    的一个法向量为···································································10

    记平面与平面夹角为

    则有··············································································12

    注:取的中点,连接易证,则为平面的一个法向量

    法二:以为坐标原点,分别为轴的正方向建系,

    可得面的一个法向量为,面的一个法向量为

    19.证:(1)由题设知:································································4

    2的所有可能取值为01235·····················································5

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

    ··················································································7

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

    ··················································································9

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

    的分布列为

    0

    1

    2

    3

    5

    P

    的数学期望为·····································································12

    20.解:(1解得·····································································3

    ∴椭圆C的方程为·····································································4

    2直线的斜率之和为0,令直线的斜率为,则直线的斜率为

    ,则的方程为

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

    ···············································································7

    则直线的斜率为,令

    同理:···········································································8

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

    同理:

    ·················································································11

    ,所以存在实数·····································································12

    方法二2直线的斜率之和为0,令直线的倾斜角为,则直线的斜率为倾斜角为··················5

    令直线的参数方程为t为参数)·························································6

    代入得:

    ·················································································7

    令点对应的参数分别为

    ·················································································9

    同理:令直线的参数方程为为参数

    代入得:

    令点对应的参数分别为·························································10

    ················································································11

    ,所以存在实数···································································12

    21.解:1,得···································································1

    由题意知:··········································································3

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

    2············································································5

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

    显然上单调递增,

    且当时,时,

    所以上递减,在上递增,

    ·················································································7

    注:这里的符号也可以通过来说明.

    故存在唯一使得···································································8

    故当时,时,

    上单调递增,在上单调递减························································9

    故存在唯一,使································································10

    故当时,时,.

    所以单调递减,在上递增·························································11

    中的较小值,又

    故恒成立.证毕·······································································12

    22.解:(1)直线l的普通方程为.·························································2

    ,则

    曲线C的直角坐标方程为································································5

    2)直线l的普通方程为t为参数)······················································6

    设点对应的参数分别为

    代入···········································································7

    ,则··········································································8

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

    23.解:1由题意得·································································3

    从而函数递减,在上递增·····························································4

    ,即.·············································································5

    21知:,又abc为正数,

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

    ·············································································7

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

    所以··············································································10

    法二:由

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

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

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