2021年高二10月月考数学试题 Word版含答案

2021年高二10月月考数学试题 Word版含答案
2021年高二10月月考数学试题 Word版含答案

xx~xx学年度

2021年高二10月月考数学试题 Word版含答案

一、选择题 (本大题共12小题,每小题4分,共48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.若直线与直线垂直,则的值为( )

A.2

B.或1

C.2或0

D.1或0

2.集合,,则( )

A. B. C. D.

3.菱形ABCD的相对顶点为,则对角线BD所在直线的方程是( )

A. B. C. D.

4.若已知函数,且,则的大小关系是( )

A. B.

C. D.

5.当圆的面积最大时,圆心坐标是( )

A. B. C. D.

6.过直线上的一点作圆的两条切线,当直线关于对称时,它们之间的夹角为( )

A.30°

B.45°

C.60°

D.90°

7设满足约束条件,若目标函数的最小值是,则的最大值为()

A.1 B.C.D.

8.已知直线,其中为实数,当这两条直线的夹角在(0,)内变动时,的取值范围是( )

A. B. C.(,1)∪(1,) D.(1,)

9.已知直线:,直线与关于直线对称,则直线的斜率为( )

A. B. C.2 D.-2

10.如果直线与圆交于M、N两点,且M、N关于直线对称,则不等式组表示的平面区域的面积是( )

A. B. C.1 D.2

11.圆心在直线上,且与两坐标轴相切的圆的标准方程为( )

A. B. 或

C. D. 或

12.方程有两个不同的解时,实数k的取值范围是( )

A. B.(,+∞) C.() D.

二、填空题(本大题共6小题,每小题4分,共24分。把答案填在题中的横线上。)

13.若满足约束条件则的最大值为__________.

14.已知,则

2

2

2)

2

2

2

2

2

+

-

x-

-

+

+的最小值为

y

+

+

-

+

+

)

x

1(

)

1(

)

1(

y

1(y

x

x

y

15.过点P()可作圆的两条切线,则的取值范围是_______.

16.已知圆,直线,下面四个结论:

①对任意实数k与θ,直线和圆M相切;

②对任意实数k与θ,直线和圆M有公共点;

③对任意实数θ,必存在实数k,使得直线和圆M相切;

④对任意实数k,必存在实数θ,使得直线和圆M相切.

其中正确结论的序号是(写出所有正确的序号)

17.已知等边△ABC的边AB所在的直线方程为,点C的坐标为(1,),则△ABC的面积为.

18.圆C经过不同的三点,已知圆C在P点处的切线斜率为1,则圆C的方程为.

三、解答题(本大题共3小题,共28分。解答应写出文字说明,证明过程或演算步骤)

19.(本小题满分8分)

已知圆C与轴相切,圆心C在直线上,且截直线的弦长为2,求圆C的方程.

20.(本小题满分10分)

过点作两条互相垂直的直线,若交轴于A点,交轴于B点,求线段AB的中点M的轨迹方程.

21.(本小题满分10分)

实系数方程的一个根在内,另一个根在内,求:

(1)的取值范围;

(2)的取值范围;

(3)的取值范围.

xx~xx学年度

高二上学期第10月考

数学试卷参考答案

一、选择题

CAABBC CCAABD

二、填空题

13.9 14. 15.16.②④17. S△ABC=

18. x2+y2+x+5y-6=0.

三、解答题

19.(本小题满分8分)已知圆C与y轴相切,圆心C在直线l1:x-3y=0上,且截直线l2:x-y=0的弦长为2,求圆C的方程.

解:∵圆心C在直线l1:x-3y=0上,

∴可设圆心为C(3t,t).

又∵圆C与y轴相切,

∴圆的半径r=|3t|…………………………………………………….4分

∴,解得t=±.

∴圆心为(,)或(-,-),半径为.

∴所求的圆的方程为(x-)2+(y-)2=或(x+)2+(y+)2=……….4分

20. 20. (本小题满分10分)过点P(2,4)作两条互相垂直的直线l1、l2,若l1交x轴于A点,l2交y 轴于B点,求线段AB的中点M的轨迹方程

解:设M的坐标为(x,y),则A、B两点的坐标分别是(2x,0)、(0,2y),连结PM,

∵l1⊥l2,

∴2|PM|=|AB|……………………………………………….5分

而|PM|=,

|AB|=

化简,得x+2y-5=0,即为所求的轨迹方程………………….5分

21.(本小题满分10分)实系数方程f(x)=x2+ax+2b=0的一个根在(0,1)内,另一个根在(1,2)内,求:

(1)的值域;

(2)(a-1)2+(b-2)2的值域;

(3)a+b-3的值域.

解:由题意…………………………………2分

易求A(-1,0)、B(-2,0).

……….2分

由∴C(-3,1).

(1)记P(1,2),k PC<

(2)|PC|2=(1+3)2+(2-1)2=17,|PA|2=(1+1)2+(2-0)2=8,|PB|2=(1+2)2+(2-0)2=13.

∴(a-1)2+(b-2)2的值域为(8,17)………………………………………….2分

(3)令u=a+b-3,即a+b=u+3.

-2

∴a+b-3的值域为(-5,-4)…………………………………………………2分30425 76D9 盙k37207 9157 酗30871 7897 碗35252 89B4 覴37122 9102 鄂.C34528 86E0 蛠25960 6568 敨27299 6AA3 檣39355 99BB 馻33550 830E 茎_

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