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(理数)09届江门市普通高中高三调研测试

(理数)09届江门市普通高中高三调研测试
(理数)09届江门市普通高中高三调研测试

江门市2009届普通高中高三调研测试

数 学(理科)

本卷共21题,满分150分,测试用时120分钟.

参考公式:如果事件A 、B 互斥,那么P (A+B )=P (A)+P (B ).

+=+n

n

a b a )(+-b a

C n 1

11

11--+n n n ab

C n

b +,其中=k

n C !

)

1()1(k k n n n +-- .

一、选择题:本大题共8小题,每小题5分,满分40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.已知全集U=R ,集合}1|{≤=y y A ,}0)2.)(1(|{>-+=x x x B ,则A B C U

)(= A .(-∞,-1] B .(2,+∞) C .[-1,1] D .φ

2.如图1,点A(2,0)、B(0,2),若复数z=a+bi (a 、b ∈R , i 为虚数单位)对应复平面内的点Z 是线段AB 的中点, 则a+b=

A .2 B.1 C .0 D .-1 3.3

2

)(x

k x -

(k 是常数)的展开式中,常数项的数值是3,则k 的值是

A .1±

B .-1

C .3±

D .-3

4.已知E 是平行四边形ABCD 的边AD 的中点,若a BA =,b BC =,则CE = A .a b 2

1+

B. a b 2

1-

C. b a 2

1+

D.b a 21-

5.一个直三棱柱,底面是等腰直角三角形,三视图 如图2,则这个直三棱柱的体积是 A .

2

2 B .2 C .

2

1 D .1

6.圆(x-1)2

+(y+1)2

=4上的点到直线3x+4y=4的最大距离与最小距离的差是 A .1 B .2 C .3 D .4

7.已知直线x+y=5和直线x+2y=8相交于点P ,在不等式组???

?

??

?≤+≤+≥≥8

2500y x y x y x 确定的平面区域内,

目标函数z=x+ay 只在点P 取得最大值,则常数a 的值可能是 A .2

3-

B.

2

3 C .2 D .1

8.已知函数m

x x x x f ++-=33)(2

3,对任意x ∈[-1,2],a x f ≤|)(|恒成立,则a 的 取值范围是 A .2

9≥

a B .a ≥5 C .a ≥7 D .a ≥9

二、填空题:本大题共7小题,你只需要作答6小题,每小题5分,满分30分.

(一)必做题(9~12题)

9.已知直线l 经过点P(-1,2),且与抛物线y=x 2

的准线 平行,则直线l 的方程是________.

10.阅读右边程序框图,若输入m=2008,则输出S=____. 11.已知一组数据:2、3、a 、b 、11、12(其中a 、b 是 常数),它们的平均数是6。若要使这组数据的标准差 最小,则常数a=____.

12.已知真命题:矩形ABCD 的对角线AC 与边AB 、AD 的

夹角分别为α、β,则1

cos cos 2

2=+βα.将这个真命题推 广,可以得到一个关于空间长方体的真命题是__________ ___________________________________________.

(二)选做题(13~15题,你只需要从中选做两题,如果三题全做,则只计算前两题的分数)

13.(坐标系与参数方程选做题)经过极点的圆ρ=4cos θ与极轴的另一个交点为P , 经过点P 的圆的切线的极坐标方程是_____. 14.(不等式选讲选做题)已知a 是常数,x=1是不等式++||a x 10||>-a x 的解,则

a 的取值范围是______. 15.(几何证明选讲选做题)如图3,⊙O 内的两条弦AB 、 CD 相交于P ,已知PA=PB=2PC=4,∠APC=600

, 则⊙O 的半径r=_____.

三、解答题:本大题共6小题,满分80分.解答须写出文字说明、证明过

程和演算步骤.

16.(本小题满分13分)已知定义在区间[0,+∞)上的函数==)(x f y )3

sin(φπ

+x A (其

中A>0,2

||π

φ<

),对0>?

x ,f(x)

(2)函数y=f(x)-2的零点从小到大依次为a 1、a 2、a 3……,求a n .

17.(本小题满分13分)一位学生每天骑车上学要经过A 、B 、C 、D 四个交通岗,每个交通岗连续循环亮绿灯、黄灯和红灯。A 、B 、D 岗一次亮绿灯、黄灯和红灯的时间都是20秒、5秒和35秒,C 岗一次亮绿灯、黄灯和红灯的时间是25秒、5秒和20秒,该学生到达每个交通岗遇到信号灯的种类和时机是随机的、相互独立。假设该学生每次上学遇到绿灯的次数为δ.

(1)求δ的分布列;

(2)求该学生上学路上遇到绿灯的平均次数.

18.(本小题满分14分)如图4,四棱锥P-ABCD 中,底面ABCD 是正方形,侧棱PD

垂直底面ABCD ,E 为PB 上一点,已知PD=AD=CD=1,PB PE 3

1=.

(1)求证:DE ⊥PB;

(2)求DE 与侧面PCD 所成角的余弦值;

(3)若过DE 且与PB 垂直的平面与BC (或其延长线) 相交于F ,求PF 的长.

19.(本小题满分14分)椭圆:

C 12

2

2

2

=+

b y

a x

)0(>>b a 的离心率2

3

=

e ,以椭圆的四个顶点为顶点的四边形的周长为54. (1)求椭圆的方程;

(2)若直线2:+=kx y l 与曲线C 恒有两个不同的交点A 和B ,求三角形OAB (其中O 为坐标原点)面积的最大值.

20.(本小题满分14分)已知函数c

bx ax x x f +++=2

ln 2)(,x>0,a 、b 、c 是常数,曲线y=f(x )在点(1,f(1))处的切线方程为y=3x+1.

(1)若x=2是函数f(x)的一个极值点,求f(x)的单调递减区间; (2)若b>1,试证明对任意x 1、x 2>0,)2(

21x x f +)]()([2

1

21x f x f +≥恒成立. 21.(本小题满分12分)数列}{n a 满足a 1=k (k 是常数),*N n ∈?,k n a a n n -+=+1. (1)若k=1,求数列}{n a 的通项:

(2)若a 2008是数列}{n a 的最小项,求k 的取值范围;

(3)若数列}{n b 对任意,n ∈N*都有++212b b +322b

11

2+-=+n n n a b ,求数列}{n b 的 前n 项和S n .

参考答案

说明:以下答案仅供参考,学生合理有效的解答均应相应给分。 一、选择题 CABD DCBA 二、填空题 9.y=2 10.

2

2009 11.4 12.参考答案:长方体ABCD-A'B'C'D'

的对角线AC'与棱AB 、AD 、AA'(或“面ABCD 、ABB'A'、ADD'A',或“面上对角线AC 、AB'、

AD'”)的夹角分别为α、β、γ,则βα22cos cos +1cos 2=+γ(或2

cos cos cos 2

22=++γβα) 13. ρcos θ=4 14.(-∞, -5)∪(5, +∞) 15.72 三、解答题

16.(1)由对0>?x ,f(x)≤f(1)=2可知:f (x)max =2 ...2分,所以A=2....3分 f(1)=2,即2)3

sin(2=+φπ

.......4分

所以1)3

sin(=+φπ,=+φπ

3

2

2ππ+k ,k ∈Z ........5分

所以6

πφ+

=k ,k ∈Z ,因为2||π

φ<

,所以6

π

φ=

? ......6分

(2)令f(x )-2=0……7分,得:1)6

3

sin(=+π

πx ....8分

=+

6

3

π

π

x 2

π+

k ,k ∈Z...9分,所以x=6k+1,k ∈Z.........10分

所以a 1=1,61=-+n n a a ,所以56-=n a n ,n ∈N* ........13分 17.(1)该学生每次到达A 、B 、D 岗遇到绿灯的概率35

520201++=p 3

1=

...1分

每次到达C 岗遇到绿灯的概率20

525252++=

p 2

1= ....2分

依题意δ=0、1、2、3、4,==)0(ξP ?-31)1(p 27

4)1(2=

-p

==)1(ξP +?-231)1(p p 2

1

113)1(p p C -??27

10)1(2=-?p

==)2(ξP ??11

3p C 2

322

1)1(C p p +?-)1(12

1p p -??31)1(2=-?p ?==2

3)3(C P ξ?-?)1(12

1p p ??+3

13

32p C p ?-0

1)1(

p 54

7)1(2=-p

==)4(ξP 54

123

1=

?p p .......7分(每正确列式并计算1个给1分)

所以,δ的分布列为

....9分

(2))0(0=?=ξξP E )1(1=?+ξP )2(2=?+ξP )3(3=?+ξP )4(4=?+ξP ...11分

2

3 =,即该学生上学路上遇到绿灯的平均次数为

2

3 .......13分

18.(1)以D 为原点,DA 、DC 、DP 为单位正交基底建立空间直角坐标系Oxyz....1分 则D(0, 0, 0), A(1, 0, 0), B(1, 1, 0), C(0, 1, 0), P(0, 0, 1) ...... 2分

)1,1,1(-=PB ,DP DE =DP PE =+=+

PB 31

)3

2

,31,31( .....4分 =

?PB DE 13

113

1?+

?0)2(3

2=-?+

.....5分,所以DE ⊥PB .....6分

(2))0,0,1(=DA 是侧面PCD 的一个法向量 ...7分 设DE 与侧面PCD 所成的角为θ,则DA <,>=

DE θπ

-2

DA =DE 6

1= ....9分,所以6

5cos =θ ...10分

(3)依题意,设F(a ,1,0),则PB ⊥EF..........11分

DF EF =)0,1,(a DE

=-)32,31,31(-)3

2

,32,31(--=a ...........12分 解=?EF PB )3

1(1-?a )1(3

21-+?

+0)3

2(=-

?,得a=-1..............13分

所以=PF 3)10()01()0(2

22=-+-+-a ..........14分

19.(1)由题意:=+224b a 5542

2=+?b a ……2分

=

=

a

c e =

-a

b a

2

2

a ?2

3

b 2= ...........4分 解得a=2,b=1......分,椭圆的方程为

14

2

2

=+y

x

……6分

(2)将2.kx y =代入14

2

2

=+y x 得22).41(x k +0428=++

kx .....7分 由直线l 与椭圆交于不同的两点得:

-=?2

)28(k 0)41(162

>+k ,即4

12

>

k

①.........8分

设A(x 1,y 1)、B(x 2,y 2),则=+21x x 2

4128k

k

+-

、2

21414k

x x +=

…9分

设直线l 与y 轴交于点D ,则:2||=OD ...10分

=-=???OAD OBD OAB S S S ?||21OD 2

22

212

2121)

14(14224)(22||+-?=-+?=-k k x x x x x x …11分

令t=4k 2+1,t>2,则4k 2=t-1

181

)411(22222222

≥+--?=-?=?t t

t S OAB

…12分 当且仅当t=4是取“=”,此时4

3.2=

k ,满足①.........13分

故三角形OAB 面积的最大值为1 ……14分

20.(1)x x f 2

)(/

=b

ax ++2 .....1分,由已知得??

???=++==++=041)2(322)1(

//

b a f b a f ....3分 解得??

?=-=3

1b a …4分,解0322

)('=+-=x x

x f )0(>x 得x=2 .......5分

当00;当x>2,f'(x )<0 ..........6分

所以f(x)的单调递减区间是[2,+∞)............7分

(2)由已知得)2(21x x f ++-)([2

1

1x f =)](2x f 212

214)(ln

x x x x +2

2.1)(4

x x a --

......9分

f'(1)=2+2a+b=3,2

1b a -=

,当b>1时a<0.....11分

由基本不等式2

1x x 221x x +≤得2

12

214)(x x x x ≥+,04)(ln 2

1

2

21≥+x

x x x ....13分 所以)2(21x x f ++-)([2111x f =)](2x f -+212

21

4)(ln x x x x 0)(4

2

21≥-x x a 恒成立..........14分 21.(1)n ≥2时,a n -a n-1=n-(k+1)

所以)(1--=n n n a a a )(2.1---+n n a a 1

12)(a a a +-++ ............1分 )1[()]1([-++-=n k n )]1(+-k )]1(2[+-++k =+k k n k n 22

1

222

++- ...2分

当n=1时,2121==k a ?+?+-k k 21212,所以对任意n ∈N *

,-=22

n a n k n k 2212++, 当k=1时,对任意n ∈N *

,22

3

22

+-=n n a n ..........4分

(2)2

2

1n

a n =

n k )212(

+-21

2=+k 2)212(?+-k n -+k 22

8

)

12(+k ,是?+=212k n 为对称轴

的二次函数 .....5分,依题意2

1

|2008212|

≤-+k ....6分,解得2007≤k ≤2008...7分 (3)n=1时,b 1=a 2=1;n ≥2时,11

2+-=n n n a b k n a n -=

-,1

2

.--=n n k n b .....8分

所以?

??

??--=1

21n k n b n

1

2=≥n n ……………9分,n=1时,S 1=b 1=1;n ≥2时, 1

2

--=

n n k n S +--+

-2

2

1n k n 122+-+

k ……10分,n n k n S 22-=+--+-121n k n 21

2

22+-+k 两式相减计算得k S n -=41

2

2--+-

n k

n ....11分,又因为--==

k S 4110

2

21k

-+,

所以,对任意n ∈N *,k S n -=4?-+--1

2

2n k n ……12分

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