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高中会考数学试卷

高中会考数学试卷
高中会考数学试卷

高中会考数学试卷

参考公式: 圆锥的侧面积公式Rl S π=圆锥侧,其中R 是圆锥的底面半径,l 是圆锥的母线长. 圆锥的体积公式S 3

1

V =

圆锥h , 其中S 是圆锥的底面面积,h 是圆锥的高. 第Ⅰ卷 (机读卷60分)

一、选择题:(共20个小题,每小题3分,共60分)

在每个小题给出的四个备选答案中,只有一个是符合题目要求的,请把所选答案前 的字母按规定要求涂抹在“机读答题卡”第1—20题的相应位置上。 1. 设全集I {0,1,2,3}=,集合{0,1,2}M =,{0,2,3}N =,则=N C M I I ( )

A .{1}

B .{2,3}

C .{0,1,2}

D .?

2. 在等比数列}{n a 中,,8,1685=-=a a 则=11a ( ) A. 4- B. 4± C. 2- D. 2±

3. 下列四个函数中,在区间(0,)+∞上是减函数的是 ( )

A .3log y x =

B .3x

y = C .12

y x =

D .1y x

=

4. 若5

4

sin =

α,且α为锐角,则αtan 的值等于 ( ) A .

5

3

B .53-

C .34

D .34-

5.在ABC ?中,,4

,2,2π

=

∠=

=A b a 则=∠B ( )

A.

3π B. 6π C. 6π或65π D. 3π或3

6. 等差数列{}n a 中,若99=S ,则=

+65a a

( )

A.0

B.1

C.2

D.3

7. 若b a c b a >∈,R 、、,则下列不等式成立的是 ( )

A.

b a 11< B.22b a > C.1

12

2+>+c b

c a D.||||c b c a > 8. 已知二次函数2()(2)1f x x =-+,那么 ( )

A .(2)(3)(0)f f f <<

B .(0)(2)(3)f f f <<

C .(0)(3)(2)f f f <<

D .(2)(0)(3)f f f <<

俯视图

9.若函数()35

1

9

1

x x f x x x +≤?=?

-+>?,则()f x 的最大值为 ( ) A .9 B .8 C .7 D .6

10.在下列命题中,正确的是 ( )

A .垂直于同一个平面的两个平面互相平行

B .垂直于同一个平面的两条直线互相平行

C .平行于同一个平面的两条直线互相平行

D .平行于同一条直线的两个平面互相平行 11.已知0x >,函数x

x y 1

+

=的最小值是 ( ) A.1 B. 2 C. 3

12. 随机调查某校50个学生在“六一”儿童节的午餐费,结果如下表:

这50( ) A.2.4,56.0 B.2.4,56.0 C.4,6.0 D.4,6.0 13. 下列命题中正确命题个数为 ( )

1?=?a b b a ○20,,?=≠?00a b a b = ○

3?=?a b b c 且,,≠≠00a b 则=a c ○4,,,≠≠≠000a b c 则()()??=??a b c a b c .1 C

14.函数x x y 2cos 2sin =是 ( )

A .周期为

2π的奇函数 B .周期为

2

π

的偶函数 C .周期为π的奇函数 D .周期为π的偶函数

15. 如图,一个空几何体的正视图(或称主视图)与侧视图(或称左视图)为全等的等边三角形,俯视图为

一个半径为1的圆,那么这个几何体的全面积为( ) A .π B .3π

C .2π

D .π+16.已知y x ,满足??

?

??≤-+≥≥.022,0,

0y x y x 则y x z +=的最大值是 ( )

B. 1

C. 2

17.以点(2,-1)为圆心且与直线0543=+-y x 相切的圆的方程为 ( )

A.3)1()2(22=++-y x

B.3)1()2(22=-++y x

C.9)1()2(22=++-y x

D.9)1()2(22=-++y x

18. 已知()3,4=a ,()2,1=-b 且()()x +⊥-a b a b ,则x 等于 ( ) A.23 B.

232 C.233 D.234

19. 要得到函数)4

2sin(π-

=x y 的图象,只要将函数x y 2sin =的图象 ( )

A .向左平移

4

π个单位; B . 向右平移

4

π个单位;C .向左平移

8

π个单位; D .向右平移

8

π个单位。

20. 猜商品的价格游戏, 观众甲:2000! 主持人:高了!

观众甲:1000! 主持人:低了! 观众甲:1500! 主持人:高了! 观众甲:1250! 主持人:低了! 观众甲:1375! 主持人:低了!

则此商品价格所在的区间是 ( ) A .(1000,1250) B .(1250,1375) C .(1375,1500) D .(1500,2000)

第Ⅱ卷 (非机读卷 共40分)

二、填空题:(本大题共4小题,每小题3分,共12分.把答案填在题中横线上) 21. 某个容量为100的样本的频率分布直方图如下,

则在区间[4,5)上的数据的频数..为 . 22. 函数()()

2log 1a f x x =-的定义域为___________.

23. 一个骰子连续投2次,点数和为4的概率

24. 阅读程序框图,若输入的n 是100,则输出的变量S= ;T= 。

输入 n S=0,T=0

n<2

S=S+n

n=n-1

T=T+n n=n-1

输出S,T

结束

开 始

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) 25.(本小题满分8分)

如图,在正四棱柱1111ABCD A B C D -中,AC 为底面 ABCD 的对角线,E 为D D 1的中点 (Ⅰ)求证:1D B AC ⊥; (Ⅱ)求证:1//D B AEC 平面.

26.(本小题满分10分) 在ABC ?中,,,A B C 为三个内角,2

()4sin sin sin 212

B

f B B B =++. (Ⅰ)若()2f B =,求角B ;

(Ⅱ)若()2f B m -<恒成立,求实数m 的取值范围.

27.(本小题满分10分)

已知函数()y f x =,*x ∈N ,*y ∈N ,满足:

① 对任意a ,*b ∈N ,a b ≠,都有()()()()af a bf b af b bf a +>+; ② 对任意*n ∈N 都有()3f f n n ??=??. (Ⅰ)试证明:()f x 为*N 上的单调增函数; (Ⅱ)求()()()1628f f f ++; (Ⅲ)令()

3n n a f =,*n ∈N ,试证明:

4

111121<+++n a a a Λ.

A1

C

参考答案

1---20

AADCB CCABB BABAB CCCDC 21、30;22、(-1,1);23、12

1

;24、2550,2500。 25、 证明:(Ⅰ)连结BD

在正四棱柱1111D C B A ABCD -中 ,ABCD 1平面⊥DD 是正方形ABCD

BD

AC ABCD AC DD AC DD ⊥∴⊥∴?⊥是正方形平面平面ΘΘ11ABCD

,ABCD

B

D AC DB D B D DB D AC D DD BD BD AC AC DD 111111,,⊥∴?⊥∴=⊥⊥平面平面ΘI Θ

(Ⅱ)设OE O,AC 连结=I BD AEC

//AEC EO ,AEC B D EO

//EO D D E DO BO ABCD 11111平面平面平面的中位线是的中点是是正方形B D B D DB D ∴??∴?∴=∴ΘΘΘ

26、解:(Ⅰ) Θf 2(B)= ∴2

1sin =

B Θπ<

56

ππ

=

∴B (Ⅱ) Θ f (B)-m<2恒成立 恒成立m 12sinB <-∴ Θπ<

27、解:(I )由①知,对任意*

,,a b a b ∈--b f a f b a ,

由于0<-b a ,从而)()(b f a f <,所以函数)(x f 为*

N 上的单调增函数.

(II )令a f =)1(,则1a …,显然1≠a ,否则1)1())1((==f f f ,与3))1((=f f 矛盾.从而1>a ,而由3))1((=f f ,即得3)(=a f . 又由(I )知a f a f =>)1()(,即3

于是得31<

a ∈N ,从而2=a ,即2)1(=f . 进而由3)(=a f 知,3)2(=f .

于是623))2(()3(=?==f f f ,

933))3(()6(=?==f f f , 1863))6(()9(=?==f f f , 2793))9(()18(=?==f f f , 54183))18(()27(=?==f f f , 81273))27(()54(=?==f f f ,

由于5427815427-=-=,

而且由(I )知,函数)(x f 为单调增函数,因此55154)28(=+=f . 从而(1)(6)(28)295566f f f ++=++=.

(III )1333))3(()(+=?==n n n n f f a f ,

n n n n a a f f f a 3))(()3(11===++,6)3(1==f a .

即数列}{n a 是以6为首项, 以3为公比的等比数列 .

∴ 16323(1,2,3)n n

n a n -=?=?=L .

于是21211(1)

111111111133()(1)1233324313

n n n

n a a a -+++=+++=?=--L L , 显然41

)3

11(41<-n ,

综上所述,

4

1

11121<+++n a a a Λ

2013全国统一高考(广西卷)数学文试题

2013年普通高等学校招生全国统一考试 数学(文科)(广西卷) 一、选择题:本大题共12小题,每小题5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)设集合{}{}1,2,3,4,5,1,2,u U A A ===集合则e (A ){}1,2 (B ){}3,4,5 (C ){}1,2,3,4,5 (D )? (2)已知a 是第二象限角,5sin ,cos 13 a a ==则 (A )1213- (B )513 - (C )513 (D )1213 (3)已知向量()()()()1,1,2,2,,=m n m n m n λλλ=+=++⊥-若则 (A )4- (B )3- (C )-2 (D )-1 (4)不等式222x -<的解集是 (A )()-1,1 (B )()-2,2 (C )()()-1,00,1U (D )()()-2,00,2U (5)()8 62x x +的展开式中的系数是 (A )28 (B )56 (C )112 (D )224 (6)函数()()()-121log 10=f x x f x x ??=+> ???的反函数 (A ) ()1021x x >- (B )()1021 x x ≠- (C )()21x x R -∈ (D )()210x x -> (7)已知数列{}n a 满足{}12430,,103 n n n a a a a ++==-则的前项和等于 (A )()-10-61-3 (B )()-1011-39 (C )()-1031-3 (D )()-1031+3 (8)已知()()1221,0,1,0,F F C F x -是椭圆的两个焦点过且垂直于轴的直线交于 A B 、两点,且3AB =,则C 的方程为 (A )2212x y += (B )22132x y += (C )22143x y += (D )22154 x y +=

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