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凤台一中2014届高三数学周练考试

一. 选择题.( 本大题共10小题，每小题5分，共50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.i 是虚数单位，

=+i

i 33( B )

（A ）

41-

（B ）i 12

341- （C ）i 6

321+ （D ）

i 6

321- 2.若集合}2

1log |{2

1≥=x x A ，则=A C R ( A )

（A ）?

??+∞?-∞,22]0,( （B ）?

??+∞,22 （C ）????

???+∞?-∞,22]0,( （D ）???

???+∞,22 3.“φ=π”是“曲线y=sin(2x ＋φ)过坐标原点的( A )

A.充分而不必要条件

B.必要而不充分条件

C.充分必要条件

D.既不充分也不必要条件 4.下列函数中，不满足：(2)2()f x f x =的是（ C ）

()A ()f x x = ()B ()f x x x =- ()C ()f x x =+1 ()D ()f x x =-

5.记cos(80)k -?=,那么tan100?=( B )

6.曲线2

y x =

+在点（-1，-1）处的切线方程为( A ) （A ）y=2x+1 (B)y=2x-1 C y=-2x-3 D.y=-2x-2

7.直线l 过抛物线C :x 2

=4y 的焦点且与y 轴垂直，则l 与C 所围成的图形的面积等于( C )

43 B.2 C.8

D.3

8. 设△ABC 的内角A , B , C 所对的边分别为a , b , c , 若cos cos sin b C c B a A +=, 则△ABC 的形状为（ B ）

(A) 锐角三角形 (B) 直角三角形

(D) 不确定

9．函数2sin 2

y x =

-的图象大致是（ C ）

10.已知函数)2sin()(?+=x x f ，其中?为实数，若|)6

(|)(π

f x f ≤对R x ∈恒成立，且

)()2

(ππ

f f >，则)(x f 的单调递增区间是( C ) (A) )(6,3

Z k k k ∈??

???+

πππ

(B) )(2,Z k k k ∈??

πππ

Z k k k ∈??

πππ

(D) )(,2Z k k k ∈??

ππ

π 二. 填空题.( 本大题共6小题，每小题5分，共30分．把答案写在答题卷相应的位置上。) 11.命题“对任何

3|4||2|,>-+-∈x x R x ”的否定是:存在

,-2-4x x x ∈≤R 使得||+| ．

12.．已知)(x f 是定义在R 上的奇函数。当0>x 时，x x x f 4)(2

-=，则不等式x x f >)( 的解集用区间表示为 (﹣5，0) ∪(5，﹢∞) ．

13.在直角坐标系xOy 中，以原点O 为极点，x 轴的正半轴为极轴建立极坐标系。若极坐标

方程为cos 4ρθ=的直线与曲线2

x t

y t

?=??=??（t 为参数）相交于,A B 两点，则______AB =16 14.若关于实数x 的不等式53x x a -++<无解，则实数a 的取值范围是_(],8-∞_______

15．已知函数32

2()(1),2x f x x x x ?≥?=??-

若关于x 的方程f(x)=k 有两个不同的实根，则数k

的取值范围是_（0，1）______

16.已知)3)(2()(++-=m x m x m x f ，22)(-=x

x g ，若同时满足条件： ①R x ∈?，0)(

则m 的取值范围是_)2,4(--∈m ______。

三. 解答题.( 本大题共5小题，共70分。解答过程应写出必要的文字说明，证明过程或演算步骤，并写在答题卷相应的位置上。) 17. （本小题满分13分）

设集合2

{|40},{|0},A x x x B x x ax a =+==++=若,A B A = 求实数a 的取值范围。

18.（本小题共13分）已知()4

3cos =

+θπ，求)cos()cos()2cos()2cos(]1)[cos(cos )cos(θθππθπθθπθθπ-+++-+-++的

值．

19.（本小题共13分）

已知函数2

()1(0),()f x ax a g x x bx =+>=+

（1）若曲线()y f x = 与曲线()y g x = 在它们的交点（1，c ）处具有公共切线，求a,b 的值；

（2）当2

a =4

b 时，求函数 ()()g x f x + 的单调区间，并求其在区间(,1-∞-]上的最大值。解：（1）由()1

c ，为公共切点可得：

2()1(0)f x ax a =+>，则()2f x ax '=，12k a =，

3()g x x bx =+，则2()=3f x x b '+，23k b =+，

∴23a b =+①

又(1)1f a =+，(1)1g b =+，

∴11a b +=+，即a b =，代入①式可得：3

3a b =??

．（2） 24a b =，∴设3221

()()()14

h x f x g x x ax a x =+=+++

则221()324h x x ax a '=++，令()0h x '=，解得：12a x =-，26

x =-；

0a >，∴26

a a -<-，

∴原函数在2a ?

-∞- ??

，单调递增，在26a a ??

-- ??

，单调递减，在6

a ??-+∞ ??

，上单调递增 ①若12a

--≤，即2a ≤时，最大值为2(1)4a h a =-；

②若126a a -<-<-，即26a <<时，最大值为12a h ??

-= ???

③若16a --

≥时，即6a ≥时，最大值为12a h ??

-= ???

．综上所述：

当(]02a ∈，时，最大值为2(1)4a h a =-；当()2,a ∈+∞时，最大值为12a h ??

-= ???

．

20．(本题满分15分)

已知函数)2

)(2

cos()sin(cos 2cos 2sin )(2

?π

?π?π?<

+---=x x x f 在

x 时取得最大值．

（Ⅰ）求?的值；

（Ⅱ）将函数)(x f y =图象上各点的横坐标扩大到原来的2倍，纵坐标不变，得到函数

)(x g y =的图象，若)0,2

(,31)(π

αα-∈=g ，求αcos 的值．

解：(Ⅰ) ???sin sin )2cos 1(cos 2sin )(++-=x x x f ………………2分

)2sin(sin 2cos cos 2sin ???-=-=x x x ………………4分

函数)(x f y =在6

x 时取得最大值

1)3sin(1)62sin(-=-?=-?∴π

??π

Z k k k ∈-

=?-=-∴,6

23π

π?π

ππ? ………………6分

又2

?π

=∴ ………………7分

（Ⅱ）由（1）可知)6

2sin()(π

=x x f ………………8分

则将函数)(x f y =图象上各点的横坐标扩大到原来的2倍，纵坐标不变，得到函数)(x g y =为)6

sin()(π

=x x g ，故3

sin()(=

=π

ααg ……10分 )0,2

(π

α-∈ )6

,3(6π

ππ

α-∈+

∴ 322)6c o s

(=+∴πα……12分

αcos 6

626sin )6sin(6cos )6cos(

]6)6cos[(+=+++=-+=ππαππαππα ………………15分

21.（本小题满分16分）设函数()1x

f x e -=-．

（Ⅰ）证明：当x ＞-1时，()1

f x x ≥+；（Ⅱ）设当0x ≥时，()1

f x ax ≤

+，求a 的取值范围．

9．函数2sin 2

y x =

-的图象大致是

（11）命题“对任何3|4||2|,>-+-∈x x R x ”的否定是．

（11）存在,-2

-4|3x x x ∈≤R 使得||+| 15、在直角坐标系xOy 中，以原点O 为极点，x 轴的正半轴为极轴建立极坐标系。若极坐

标方程为cos 4ρθ=的直线与曲线2

x t

y t

?=??=??（t 为参数）相交于,A B 两点，则______AB = 【答案】：16

16、若关于实数x 的不等式53x x a -++<无解，则实数a 的取值范围是_________ 【答案】：(],8-∞

16．本小题主要考查三角函数的基本公式和解斜三角形的基础知识，同时考查基本运算能力。

（满分10分）解：（Ⅰ）2

2cos 14444

c a b ab C =+-=+-?

= 2.c ∴=

ABC ∴?的周长为122 5.a b c ++=++=

（Ⅱ）1cos ,sin 44

C C =

∴===

sin 4sin 28

a C A c ∴===

a c A C

<∴<

，故A为锐角，

cos.

∴===

7111

cos()cos cos sin sin.

848816

A C A C A C

∴-=+=?+?=

（16）(本小题满分12分)

设函数2

()cos(2)sin

f x x x

=++

（I）求函数()

f x的最小正周期；

（II）设函数()

g x对任意x R

∈，有()()

g x g x

+=，且当[0,]

∈时，

()()

g x f x

=-；

求函数()

g x在[,0]

π-上的解析式。

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