浙江省湖州中学2014届高三上学期期中考试数学理试题 Word版含答案

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试 卷

一、选择题:本大题共10小题,每小题5分,共50分. 1.全集{0,1,2,3}U =,{2}U C M =,则集合M =( )

A .{0,1,3}

B .{1,3}

C .{0,3}

D .{2}

2.若函数()f x (x R ∈)是奇函数,函数()g x (x R ∈)是偶函数,则( )

A .函数()()f x g x +是奇函数

B .函数()()f x g x ?是奇函数

C .函数[()]f g x 是奇函数

D . 函数[()]g f x 是奇函数 3. 下列函数中,图像的一部分如右图所示的是( ) A .sin()6

y x π

=+

B .sin(2)6

y x π

=-

C .cos(2)6y x π

=-

D .cos(4)3

y x π

=-

4.等比数列{a n }的前n 项和为S n ,若S 2n =3(a 1+a 3+…+a 2n -1),a 1a 2a 3=8,则a 10等于( )

A .-1024

B .1024

C .-512

D .512

5.已知函数2

()f x x bx =+的图象在点A (1,f (1))处的切线的斜率为3,数列1()f n ??

????

的前n 项

和为n S ,则2013S 的值为( )

A .

2010

2011

B .

20112012 C .2012

2013

D .

2013

2014

6.若实数x ,y 满足不等式组2402300x y x y x y +-≥--≥-≥??

???

, 则x +y 的最小值是( )

A .43

B .3

C . 4

D . 6

第3题图

8.命题p :“1≠x 或3y ≠”是命题q :“4≠+y x ”的( )条件

A .充分不必要

B .必要不充分

C .充要

D .既不充分也不必要

9.如图,半圆的直径6AB =,O 为圆心,C 为半圆上不同于A 、B 的任意一点,若P 为半

径OC 上的动点,则()

PA PB PC +的最小值为( )

A .92

B .9

C .92

- D .-9

10.若函数3

2

()f x x ax bx c =+++有两个极值点12,x x ,且11()f x x =,则关于x 的方程

23(())2()0f x af x b ++=的不同实根个数是( )

A .3

B .4

C .5

D .6

二、填空题:本大题共7小题,每小题4分,共28分. 11. 不等式2

(2)211x x -≤+的解集为 ____.

12. 已知数列{}n a 满足11a =,12n n n a a +=+,则10a =_________. 13.在ABC ?中,,,a b c 分别是内角,,A B C 的对边,已知1

6,4,cos 3

a c B ===,则____

b =. 14. 已知41)4

sin(=

+

π

θ, ),23(ππθ--∈,则)127cos(πθ+的值为________.

15. 若)4

sin(3)4

sin()(π

π

-

++=x x a x f 是偶函数,则=a .

16. 函数2

1()2ln 2

f x x x x a =+-+在区间(0,2)上恰有一个零点,则实数a 的取值范围是_____.

17. 已知函数2

()|21|f x x x =+-,若1a b <<-且()()f a f b =,则ab a b ++的取值范围_____.

O

P C B

A

浙江省湖州中学

2013学年第一学期高三期中考试

数 学 答 卷(理)

一、选择题:本大题共10小题,每小题5分,共50分.

二、填空题:本大题共7小题,每小题4分,共28分.

11___________________ 12_________________ 13______________________

14___________________ 15_________________ 16______________________

17___________________

三、解答题:第18、19、20题每题14分,第21、22每题15分,共72分.

18.已知 1:(),3

x

p f x -=且|()|2f a <;

:q 集合{

}

2(2)10,A x x a x x R =+++=∈,{}0B x x =>且 A B =?.

若p ∨q 为真命题,p ∧q 为假命题,求实数a 的取值范围.

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班级 学号______ 姓名 试

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19.已知函数2()sin

2cos 24

x x f x =+. (1)写出如何由函数sin y x =的图像变换得到()f x 的图像;

(2)在ABC ?中,角A B C 、、所对的边分别是a b c 、、,若C b B c a cos cos )2(=-,求

)(A f 的取值范围.

20.已知函数2

()(x

x

f x a x a =+∈R ,1)a >, (1)求函数f (x )的值域;

(2)记函数()(),[2,)g x f x x =-∈-+∞,若()g x 的最小值与a 无关,求a 的取值范围;

(3)若m >,直接写出(不需给出演算步骤)关于x 的方程()f x m =的解集.

21.已知数列{}n a 的前n 项和11()22

n n n S a -=--+(n 为正整数).

(1)令2n n n b a =,求证数列{}n b 是等差数列,并求数列{}n a 的通项公式;

(2)令1n n n c a n +=,12 n n T c c c =+++,试比较n T 与

521

n

n +的大小,并予以证明.

·································································································································· ··································································································································

班级 学号______ 姓 试

·································································································································· ·································································································································· ----------------------------------------装-----------------------------------------订----------------------------------线-------------------

22.已知实数a 满足02a <≤,1a ≠,设函数3211()32

a f x x x ax +=-+. (1)当2a =时,求()f x 的极小值;

(2)若函数32

()(24)ln g x x bx b x x =+-++(b R ∈)的极小值点与()f x 的极小值点相同.求证:()g x 的极大值小于等于

54

浙江省湖州中学

2013学年第一学期高三期中考试

数 学 答 卷(理)

一、选择题:本大题共10小题,每小题5分,共50分.

二、填空题:本大题共7小题,每小题4分,共28分.

11___[1,7]-_______ 12____1023______ 13___________6_____

14_______ 15____3-_____ 16____2ln 24a ≤-或32a =-__

17_______(1,1)-____

三、解答题:第18、19、20题每题14分,第21、22每题15分,共72分.

18.已知 1:(),3

x

p f x -=且|()|2f a <;

:q 集合{

}

2(2)10,A x x a x x R =+++=∈,{}0B x x =>且 A B =?.

若p ∨q 为真命题,p ∧q 为假命题,求实数a 的取值范围.

解:对p :所以1|()| |

|23

a

f a -=<.若命题p 为真,则有75<<-a ;...........2分

q :∵}0x |x {B >=且 ?=?B A

∴若命题q 为真,则方程01x )2a (x )x (g 2

=+++=无解或只有非正根.

∴04)2a (2

<-+=?或0(0)0202

g a ?

??≥?≥??+?-...........................5分

∵p, q 中有且只有一个为真命题

∴ (1) p 真,q 假:则有4a 54a 7

a 5-≤<-?

?

?-≤<<-,即有;......................8分 ·································································································································· ··································································································································

班 学号______ 姓名 试

·································································································································· ·································································································································· ----------------------------------------装-----------------------------------------订----------------------------------线-----------------

(2) p 假,q 真:则有7a 4a 5

a 7a ≥?

?

?->-≤≥,即有或;

∴4a 5-≤<-或7a ≥. ........................14分

19.已知函数2()sin

2cos 24

x x f x =+. (1)写出如何由函数sin y x =的图像变换得到()f x 的图像;

(2)在ABC ?中,角A B C 、、所对的边分别是a b c 、、,若C b B c a cos cos )2(=-,求

)(A f 的取值范围.

解:()142sin 212cos 2sin

+??

?

??+=++=πx x x x f ……………………3分 (Ⅰ) 24sin sin()sin()424

x y x y x y π

π

π=????→=+?????→=+左移

横坐标伸长为

原来的倍个单位

1sin()sin()12424

x x y y ππ→=+???→=++上移

个单位 ……7分 (Ⅱ)由()C b B c a cos cos 2=-,利用三角形中的正弦定理知:1cos 2=B ∵π<

π

=

B ……………………10分

()142sin 2+??

?

??+=πA A f ,

∵320π<

7424π

ππ<+

142sin 22≤??

?

??+<πA ,……………………12分 ∴()122+≤

20.已知函数2

()(x

x

f x a x a =+

∈R ,1)a >, ················ ············

····

············

···· ················ ----

(1)求函数f (x )的值域;

(2)记函数()(),[2,)g x f x x =-∈-+∞,若()g x 的最小值与a 无关,求a 的取值范围;

(3)若m >,直接写出(不需给出演算步骤)关于x 的方程()f x m =的解集.

解:(1)①0x ≥时,221,()x

x x

x x

a f x a a a a ≥=+=+≥ ,

当且仅当2x x a a

=

,即1x

a =>时等号成立; ②0x <,3

1,01,()3x x a a f x a

>∴<<∴=> ,

由①②知函数()f x 的值域为)+∞.

(2)()()2,[2,)x

x g x f x a a x =-=+∈-+∞, ①0x ≥,1,1,()3,()3x

x

a a g x a g x >∴≥=∴≥ ,

②20x -≤<时,2

11,

1,()2x x x

a a g x a a a ->≤<=+ , 令x t a =,则1()2g x t t =+,记1()2.h t t t =+21

(1)t a

≤<,

1

()2h t t t =+≥1

2t t

=,t =

(i)21a ≤,即a ≥时,结合①知min ()g x =与a 无关;

(ii)21a >,即1a <<时,421()220h t a t

'=-≥->, ()h t ∴在21[,1)a 上是增函数,2min min 2212

()()()3g x h t h a a a

===+<,

结合①知2min 22

()g x a a

=+与a 有关;

综上,若()g x 的最小值与a 无关,则实数a 的取值范围是a ≥

(3)①3m <≤时,关于x 的方程()f x m =的解集为|log a x x ??=???;

②m >3时,关于x 的方程()f x m =的解集为|log a x x ??

=???

3log a x m ?=??.

21.已知数列{}n a 的前n 项和11

()22

n n n S a -=--+(n 为正整数).

(1)令2n n n b a =,求证数列{}n b 是等差数列,并求数列{}n a 的通项公式;

(2)令1n n n c a n +=

,12 n n T c c c =+++,试比较n T 与

521

n

n +的大小,并予以证明. 解(I )在11()22n n n S a -=--+中,令n=1,可得1112n S a a =--+=,即11

2

a =

当2n ≥时,21111111

()2()22

n n n n n n n n n S a a S S a a ------=--+∴=-=-++,,

11n 111

2a (),212

n n n n n a a a ----∴=+=+n 即2.

112,1,n 21n n n n n n b a b b b --=∴=+≥-= n 即当时,b . 又1121,b a ==∴数列}{

n b 是首项和公差均为1的等差数列. 于是1(1)12,2n n n n n

n b n n a a =+-?==∴=. (II)由(I )得11

(1)()2

n n n n c a n n +=

=+,所以 231111

23()4()(1)()2222

n n T n =?+?+?+++K

2341111112()3()4()(1)()22222

n n T n +=?+?+?+++K 由①-②得23111111

1()()()(1)()22222n n n T n +=++++-+K 111

11[1()]

133421(1)()1222123

32

n n n n n

n n n T -++-+=+-+=--+∴=- 535(3)(221)3212212(21)

n n n n

n n n n n T n n n ++---=--=+++ 于是确定521

n n T n +与的大小关系等价于比较221n

n +与的大小 由

可猜想当322 1.n

n n ≥>+时,

证明如下: 证法1:(1)当n=3时,由上验算显示成立。

22.已知实数a 满足02a <≤,1a ≠,设函数3211()32

a f x x x ax +=-+. (1)当2a =时,求()f x 的极小值;

(2)若函数3

2

()(24)ln g x x bx b x x =+-++(b R ∈)的极小值点与()f x 的极小值点相同.求证:()g x 的极大值小于等于

5

4

(Ⅰ) 解: 当a =2时,f ′(x )=x 2-3x +2=(x -1)(x -2). `列表如下:

x (-∞,1) 1

(1,2) 2 (2,+∞) f ′(x ) + 0 - 0 +

f (x ) 单调递增 极大值 单调递减 极小值

单调递增

所以,f (x )极小值为f (2)=2

3

. …………………………………5分

(Ⅱ) 解:f ′(x )=x 2-(a +1)x +a =(x -1)(x -a ).

g ′(x )=3x 2

+2bx -(2b +4)+

1x

2(1)[3(23)1]

x x b x x

-++-.

令p (x )=3x 2+(2b +3)x -1, (1) 当 1<a ≤2时,

f (x )的极小值点x =a ,则

g (x )的极小值点也为x =a , 所以p A .=0, 即3a 2+(2b +3)a -1=0,

即b=

2

133

2

a a

a

--

此时g(x)极大值=g(1)=1+b-(2b+4)=-3-b

=-3+

2

331

2

a a

a

+-

313

222

a

a

--.

由于1<a≤2,

故313

222

a

a

--≤

3

2

?2-

1

4

3

2

5

4

.………………………………10分

(2)当0<a<1时,

f (x)的极小值点x=1,则g(x)的极小值点为x=1,由于p(x)=0有一正一负两实根,不妨设x2<0<x1,所以0<x1<1,

即p(1)=3+2b+3-1>0,

故b>-5

2

此时g(x)的极大值点x=x1,

有g(x1)=x13+bx12-(2b+4)x1+ln x1

<1+bx12-(2b+4)x1

=(x12-2x1)b-4x1+1(x12-2x1<0)

<-5

2

(x12-2x1)-4x1+1

=-5

2

x12+x1+1

=-5

2

(x1-

1

5

)2+1+

1

10

(0<x1<1)

≤11

10

5

4

综上所述,g(x)的极大值小于等于5

4

.……………………14分

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