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武汉市部分重点中学2017—2018高一上学期期末数学试卷(五校联考)含详细答案

武汉市部分重点中学2017—2018高一上学期期末数学试卷(五校联考)含详细答案
武汉市部分重点中学2017—2018高一上学期期末数学试卷(五校联考)含详细答案

武汉市部分重点中学2017—2018学年度上学期期末测试

高一数学试卷

命题人:武汉市第14中学

一、选择题(本大题共12小题,每小题5分,共60分,每小题给出的四个选项中,只有一项是正确的)

1.已知,,那么的终边所在的象限为()

A. 第一象限

B.第二象限

C.第三象限

D.第四象限

2.已知集合,,则=()

A B C D

3.已知函数若对任意的都有,则

=()

A. 0

B. -3

C. 3

D. 以上都不对

4.已知,则=()

A B C D

5.已知函数则函数的零点的个数为()

A. 1

B. 2

C. 3

D. 4

6.计算等于()

A. B. C. D.

7.=()

A. B. C. D.

8.已知,且,则的值为()

A. B. C. D.

9.已知函数,的最小正周期为,则函数的图像的一条对称轴方程是()

A. B. C. D.

10.下列结果为的是()

①;

②;③;④

A.①②

B. ①②③

C. ①③④

D. ①②③④

11.对于函数,,有以下四个判断:

①把的图像先沿x轴向左平移个单位,再将纵坐标伸长到原来的4倍(横坐标不变)后就可以等到函数图像;②该函数图像关于点对称;③由可得必是的整数倍;④函数在上单调递增。其中正确判断的个数是()

A. 1

B. 2

C. 3

D. 4

12.设,且满足,则x+y=()

A. 0

B. 1

C. 2

D. 4

二、填空题(本大题共4小题,每小题5分,共20分)

13.求使不等式,成立的x的取值集合为

14.化简:=

15.若,在区间上的最大值为1,则

16.对实数和,定义运算“”:,设函数,

,下列说法正确的序号是

①是周期函数,且周期为;

②该函数的值域为;

③该函数在上单调递减;

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.(本小题满分10分)已知集合,函数

,求时的最大值。

18.(本小题满分12分)已知,,且、是方程

的两个根,求的值。

19.(本小题满分12分)某同学用“五点法”画函数,

在某一个周期内的图像时,列表并填入了部分数据,如下表:

(I)请求出上表中的,并直接写出函数的解析式;

(II)将的图像上点的横坐标变为原来的2倍,纵坐标不变;再把所得图像上所有点向右平移个单位,得到函数,若当且时,求的值。

20.(本小题满分12分)定义运算:,设函数

函数

(I)求函数的最小正周期

(II)若函数在上的最大值与最小值之和为,求实数的值

21.(本小题满分12分)人口问题是当今世界各国普遍关注的问题,1989年7月11日本定为“世界人口日”,以引起国际社会对人口问题的重视

(I)世界人口在过去的50年里翻了一番,问每年世界人口平均增长率是多少?(II)我国人口在1998年底达到12.48亿,若将人口平均增长率控制在1%以内,我国人口在今年(2018年)底至多将有多少亿?

以下数据供计算时使用

22.(本小题满分12分)对定义在上,并且同时满足以下两个条件的函数

称为“漂亮函数”

①对任意的,总有

②当时,总有成立

已知函数与和都是定义在上的函数(I)请证明函数为“漂亮函数”;

(II)试问函数是否为“漂亮函数”?并说明理由;

(III)若函数是“漂亮函数”,求实数b组成的集合。

参考答案及解析

13.14. 1 15. 16. ①②③④详细解析如下:

一、选择题

1.D

∵∴∴

故在第四象限

2.B

3.A

∵∴对称轴为

=

=

=0

4.B

5.D

由图像可知,的零点个数必有4个

6.C

∵∴

原式=

=

=

7.C

原式=

=

=

=

8.D

∵∴

∵且∴

原式=

=

=

9.B

=

=

∵∴∴

对称轴

k=0时,

10.C

=

=

=

=

=

=

=

=

④原式=

11.B

①向左平移个单位得即,横坐标不变,纵坐标伸长4倍得,故①错误

②正确

③∵且

∴相隔的整数倍,故错误

为最大值

故④正确

12.B

∴两式相加为0

设,显然为奇函数且单调递增

∴,即

13

14.原式=

=

=

=1

15. ∵

∴在上单调递增

∵在最大值为1

16.画出函数图像,由图像可知,,故①②③④都正确

17.∵

在上单调递增∴

18.∵

∵,

∴,

19.(I)解得

B=0

∴解得

(II)=

∵∴∴∵∴∴

=

=

=

20.(I)

=

=

=

(II)=

∴,

21.(I)设平均增长率为x

(II)

∴N=1.218

22.(I)∵在上单调递增

即时总有

当时

=

=

∴为“漂亮函数”

(II)在上单调递增∴0

=

=

不是“漂亮函数”(III)∵

=

当时,∴

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————————————————————————————————作者:————————————————————————————————日期:

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