高中会考数学考试试题

高中会考数学考试试题
高中会考数学考试试题

2011级高中数学毕业会考试题

命题: 二高高二数学组 2012.11.10

一、选择题(共20个小题,每小题3分,共60分)每题只有一个符合题目要求,请把所选答案涂在“机读答题卡”相应位置上

1.已知集合{}{}13,25A x x B x x A B =-≤<=<≤=,则( ) A. ( 2, 3 ) B. [-1,5] C. (-1,5) D. (-1,5] 2.sin

3π4cos 6π5tan ??

?

??3π4-=( ).A .-433 B .433

C .-

43

D .4

3 3.奇函数)(x f 在区间[]a b --,上单调递减,且)0(0)(b a x f <<>,那么)(x f 在区间[]b a ,上( )

A .单调递减

B .单调递增

C .先增后减

D .先减后增

4.盛有水的圆柱形容器的内壁底面半径为5,两个直径为5的玻璃小球都浸没于水中,若取出这两个小球,

则水面将下降的高度为( )A 、53 B 、3 C 、2 D 、 4

3

5.已知关于某设备的使用年限x 与所支出的维修费用y(元)有如下表统计资料:若y 对x 呈线性相关关系,则回归直线方程y bx a =+表示的直线一定过定点( ) A (3,4) B (4,6) C (4,5) D (5,7) 6.在等比数列{}n a 中,若32a =,则12345a a a a a = ( ) (A )8

(B )16

(C )32

(D )

7.在某次测量中得到的A 样本数据如下:82,84,84,86,86,86,88,88,88,88.若B 样本数据恰好是A 样本数据

都加2后所得数据,则A ,B 两样本的下列数字特征对应相同的是( ) A .众数 B .平均数 C .中位数 D .标准差

8.已知点()0,0O 与点()0,2A 分别在直线y x m =+的两侧,那么m 的取值范围是 ( )

(A )20m -<< (B )02m << (C )0m <或2m > (D )0m >或2m <-

9.函数sin 26y x π??

=+ ??

?

图像的一个对称中心是 ( )

(A )(,0)12

π

-

(B )(,0)6

π

-

(C )(,0)6

π

(D )(,0)3

π

10.已知0a >且1a ≠,且23a a >,那么函数()x f x a =的图像可能是( )

(A ) (B ) (C ) (D )

A '

G

F

E

D

C B

A

11.ABCD 为长方形,AB =2,BC =1,O 为AB 的中点,在长方形ABCD 内随机取一点,取到的点到O 的距离大

于1的概率为( )

A. π4 B .1-π4 C.π8 D .1-π

8

12.某几何体的正视图和侧视图均如图1所示,则该几何体的俯视图不可能是( )

13.有四个幂函数:①()1

f x x -=; ②()2

f x x -=; ③()3

f x x =; ④()1

3

f x x =.某同学研究了其中的一个

函数,他给出这个函数的两个性质:(1)定义域是{x | x ∈R ,且x ≠0};

(2)值域是{y | y ∈R ,且y ≠0}.如果这个同学给出的两个性质都是正确的,那么他研究的函数是 ( )

(A )① (B )② (C )③ (D )④

14.如图,正ABC ?的中线AF 与中位线DE 相交于G ,已知ED A '?是AED ?绕DE 旋转过程中的一个图形, 下列四个命题正确的个数为( )

①动点'A 在平面ABC 上的射影在线段AF 上; ②恒有平面BCED GF A 平面⊥'; ③三棱锥FED A -‘

的体积有最大值; ④异面直线E A ’与BD 不可能垂直.

A 3

B 1

C 2

D 4

15.把函数y =sin x (x ∈R )的图象上所有点向左平行移动3

π

个单位长度,再

把所得图象上所有点的横坐标缩短到原来的2

1

倍(纵坐标不变),得到的图象所

表示的函数是( ).

A .y =sin ??

? ?

?3π - 2x ,x ∈R B .y =sin ??

? ??6π + 2x ,x ∈R C .y =sin ??

? ?

?3π + 2x ,x ∈R D .y =sin ??

? ?

?32π + 2x ,x ∈R

16.有5件产品.其中有3件一级品和2件二级品.从中任取两件,则以0.7为概率的是( )

A 至多有1件一级品

B .恰有l 件一级品

C .至少有1件一级品

D .都不是一级品 17.△ABC 中,45A ∠=?,105B ∠=?,A ∠的对边2a =,则C ∠的对边c 等于 ( ) (A )2 (B

(C

(D )1

A 图1

B C D

18.如果执行右面的程序框图,那么输出的S 等于( )

(A )45 (B )55 (C )90 (D )110 19.已知直线420mx y +-=与250x y n -+=互相垂直,垂足为

()1,p p ,则m n p -+的值是( ) A .24 B .20 C . 0 D .-4

20.如果方程x 2-4ax +3a 2=0的一根小于1,另一根大于1,那么实数a 的取值范围是 ( )

(A )1

13

a << (B )1a > (C )13

a < (D )1a = 二、填空题(共4道小题,每小题3分,共12分)

21.某校有高级教师26人,中级教师104人,其他教师若干.为

了解该校教师的工资收入情况,若按分层抽样从该校的所有教师中抽取56人进行调查,已知从其他教师中共抽取了16人,则该校共有教师 人.

22.直线b x y +=与曲线21y x -=有两个交点,则b 的取值 范围是 ;

232,()()

22-=-?+b a b a ,则a 与b 的夹角为 24.16.下列说法中正确的有_______

①刻画一组数据集中趋势的统计量有极差、方差、标准差等;刻画一组数据离散程度统计量有平均数、中位数、众数等。②抛掷两枚硬币,出现“两枚都是正面朝上”、“两枚都是反面朝上”、“恰好一枚硬币正面朝上”的概率一样大,③有10个阄,其中一个代表奖品,10个人按顺序依次抓阄来决定奖品的归属,则摸奖的顺序对中奖率没有影响。④向一个圆面内随机地投一个点,如果该点落在圆内任意一点都是等可能的,则该随机试验的数学模型是古典概型。

2011级高中数学毕业会考答题页(2012.12)

班级 学籍号 姓名 成绩

21. 22. 23. 24. . 三、解答题(共3道小题,共28分)

25.( 8分) 如图所示,在四棱锥P —ABCD 中,底面为直角梯形,AD ∥BC ,∠BAD=90°,PA ⊥底面ABCD ,且PA=AD=AB=2BC ,M 、N 分别为PC 、PB 的中点. (1)求证:PB ⊥DM ;

(2)求BD 与平面ADMN 所成的角.

26.( 10分)在△ABC 中,设A 、B 、C 的对边分别为a 、b 、c ,向量m =(cosA,sinA),n =(

2

-sinA,cosA),

若|m +n |=2. (1)求角A 的大小; (2)若b=42

,且c=

2

a ,求△ABC 的面积.

27.( 10分)已知等差数列}{n a 的前n 项和为S n ,且2

62-+=n n S n (*N n ∈),

(1)求数列}{n a 的通项公式a n ; (2)设n

a a

b n n n +=

+11,求数列{b n }的前n 项和T n

2011级高中数学毕业会考答案 2012.11.10

一、选择题:(共20个小题,每小题3分,共60分) BABAC |CDBAA |BDAAB |CACBA

二、填空题:(4个小题,每题3分,共12分)21. 182 22. ]1,2(--;23. 。60 24. _ ③ 三、解答题(共3道小题,共28分)

25.(8分)在△ABC 中,设A 、B 、C 的对边分别为a 、b 、c ,向量m =(cosA,sinA),n =(2-sinA,cosA),若|m +n |=2.

(1)求角A 的大小; (2)若b=42

,且c=

2

a ,求△ABC 的面积.

解 (1)m +n =(2

+cosA-sinA,cosA+sinA), |m +n |2=(

2

+cosA-sinA)2+(cosA+sinA)2

=2+2

2

(cosA-sinA)+(cosA-sinA)2+(cosA+sinA)2=2+2

2

(cosA-sinA)+2

=4-4sin (A-4

π)∵|m +n |=2,∴4-4sin (A-4

π)=4,sin (A-4

π)=0. 又∵0<A <π,∴-4

π<A-4

π<4

3π,∴A-4

π=0,∴A=4

π.

(2)由余弦定理,a 2=b 2+c 2-2bccosA,又b=42

,c=

2

a,A=4

π,得a 2=32+2a 2-2×4

2

×

2

a ·

2

2

,

即a 2-8

2

a+32=0,解得a=4

2

,∴c=8.∴S △ABC =2

1b ·csinA=2

1×4

2

×8×sin 4

π=16.

S △ABC =2

1×(4

2

)2=16.

26.(10分) (1)证明 ∵N 是PB 的中点,PA=PB ,∴AN ⊥PB.∵∠BAD=90°,∴AD ⊥AB.∵PA ⊥平面ABCD ,∴PA ⊥AD.∵PA ∩AB=A ,∴AD ⊥平面PAB ,∴AD ⊥PB. 又∵AD ∩AN=A ,∴PB ⊥平面ADMN. ∵DM ?平面ADMN ,∴PB ⊥DM.

(2)解 连接DN ,∵PB ⊥平面ADMN ,∴∠BDN 是BD 与平面ADMN 所成的角,

在Rt △BDN 中,sin ∠BDN=BD BN =AB

AB

2221

?=21

, ∴∠BDN=30°,

即BD 与平面ADMN 所成的角为30°.

27.( 10分)已知等差数列}{n a 的前n 项和为S n ,且2

6

2-+=n n S n (*N n ∈),

(1)求数列}{n a 的通项公式a n ; (2)设n

a a

b n n n +=

+11,求数列{b n }的前n 项和T n

解:(Ⅰ)由题意,当n=1时,a 1=S 1=-2 当2≥n 时,有

.26

)1()1(26221

n n n n n S S a n n n =--+---+=-=- ∴?

??≥=-=)2( )1(2n n n a n …………1分

(Ⅱ)当n=1时,3

1

122111211-=+?-=+=a a b 当2≥n 时, )

2(1

21)1(1121+=

+=++=

+=

=n n n n n n n n

a a

b n n n

)211(21+-=n n ∴数列{b n }的前n 项和

T n =b 1+b 2+…+b n =)

2

1111116

14

15

13

14

12

1(213

1+-++--+

+-+-+-+-n n n n =1111111111()()322

3

1

2

12

21

2

n n n n -++--=-+++++

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————————————————————————————————作者:————————————————————————————————日期: 2

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