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黑龙江省大庆试验中学09-10学年高一下学期期中考试(数学)

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大庆实验中学 2009--2010 学年度下学期期中考试
高一年级数学试题

一、选择题(本大题共 12 小题,每题 5 分,共 60 分)

1. 1 sin ? ? 3 cos ? 的值为 2 12 2 12

()

A. 0

B. ? 2 2

C. 2

D. 2

2.在 △ABC 中, AB ? 3 , A ? 45 , C ? 60? ,则 BC ?

()

A. 3 ? 3

B. 2 C. 2

3.设等比数列{ an }的前 n 项和为 Sn ,若

A.2

B. 7 3

C. 8 3

D. 3 ? 3

S6 =3 ,则 S9 的值是

S3

S 6

D.3

()

4. 在 ?ABC 中,若 a cos A ? bcosB ,则 ?ABC 是

()

A.等腰三角形 B.直角三角形 C.等边三角形 D.等腰三角形或直角三角形

5.在各项均为正数的等比数列?an? 中,若 a2a4 ? a3a6 ? a4a5 ? a5a7 ? 36 ,则

a3 ? a6 ?
A.9

B.4

C.6

D.12

()

6.函数 y ? sin(2x ? ? ) ? sin 2x 的一个单调增区间是 3

()

A.

????

? 6

,? 3

? ??

B.

?? ??12

,7? 12

? ??

C.

? ??

5? 12

,13? 12

? ??

D.

?? ?? 3

,5? 6

? ??

7.在 △ABC 中, c ? 2,b ? 6,B ? 120 ,则 a 等于

()

A. 6

B.2

C. 3

D. 2

8.在等差数列 ?an ?中, 3(a3 ? a5 ) ? 2(a7 ? a10 ? a13 ) ? 24 ,则前 13 项之和 S13 等于(

)

A.26

B.13

C.52

D.156

9.若 3sin? ? cos? ? 0 ,则

1

的值是

cos2 ? ? sin 2?

()

A. 10 3

B. 5 3

C. 2 3

D. ? 2

10.数列1,1? 2 ,1? 2 ? 22 ,1? 2 ? 22 ? 23 ,?1? 2 ? 22 ? ?? 2n?1 ,?的前 n 项和 Sn ? 1020 ,那么 n

的最小值是

()

A.7

B.8

C.9

D.10

数学试题 第 1 页 共 2 页

11.已知等差数列 ?an ?中, S n 是前 n 项和,若 S16 ? 0 且 S17 ? 0 ,则当 S n 最大时, n 的值为( )

A.16 B.9

C.8

D.10

12.设 a ? 0 , b ? 0 ,若 3 是 3a 与 3b 的等比中项,则 1 ? 1 的最小值为 ab

()

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A.8

B.4

C.1

D. 1 4

二、填空题 (本大题共 4 小题,每小题 5 分,共 20 分)

13. tan15 ? 1 ? tan 2 15 ?

? __________。

14.已知关于 x 的不等式 ax2 ? 2x ? c ? 0 的解集为 (? 1 , 1) ,则 ?cx2 ? 2x ? a ? 0 的解集为________________。 32

15.在锐角 ?ABC 中, BC ? 1, B ? 2A, 则 AC 的取值范围是



16.已知 cos(? ? x) ? 3 , 17? ? x ? 7? ,则 sin 2x ? 2sin 2 x 的值是_________。

4

5 12

4

1 ? tan x

三、解答题(本大题共 6 小题,共 70 分)

17.(本小题 10 分)解关于 x 的一元二次不等式 x2 ? (3 ? a)x ? 3a ? 0 。

18.(本小题 12 分)若 f (x) ?

3

sin

2x

?

2 cos2

x

?

m 在区间

???0,?2

? ??

上的最大值是

6,

(Ⅰ)求常数 m 的值;

(Ⅱ)求 f (x) 当 x ? R 时的最小值及相应的 x 的取值集合。

? ? 19.(本小题
20.(本小题

12 12

分)已知数列?an? 的前 n 项之和 Sn ? n2 ? 4n ? 1 ,求数列 an 的前
分)在 △ABC 中,内角 A,B,C 的对边分别是 a,b,c ,已知 c ?

n 2

项和
,C

Tn ?


?。

3

(Ⅰ)若 △ABC 的面积等于 3 ,求 a,b ;

(Ⅱ)若 sin C ? sin(B ? A) ? 2sin 2A ,求△ABC 的面积。

21.(本小题 12 分)某个体户,一月初向银行贷款 1 万元作为开店启动资金,每月月底获得的利润是该月月初投

入资金的 20%,每月月底需要交纳所得税为该月利润的 10%,每月的生活费开支为 540 元,余额作为资金全部投

入下个月的经营,如此不断继续,问到这年年底该个体户还贷款前尚余多少资金?若银行贷款的年利息为 5%,

问该个体户在年底还清银行贷款后还有多少资金?(参考数据:1.1810 ? 5.23,1.1811 ? 6.18,1.1812 ? 7.29.结果精

确到 0.1 元)

22.(本小题 12 分)已知 a1 ? 9 ,点 (an, an?1) 在函数 f (x) ? x2 ? 2x 的图象上,其中 n ? N ? ,设 bn ? lg(1? an ) .

(Ⅰ) 证明数列?bn? 是等比数列;

(Ⅱ) 设 cn ? nbn ,求数列 ?cn ?的前 n 项和 S n ;

? ? (Ⅲ)

设 dn

?

1 an

?

1 an ? 2

,求数列

dn

的前 n 项和 Dn 。

数学试题 第 2 页 共 2 页

参考答案(高考学习网)

一、选择题: BBBDC BDAAD CB 二、填空题

13. 3 14. (?2,3) 6
三、解答题

15. ( 2, 3) 16. ? 28 75

17.解:∵ x2 ? (3 ? a)x ? 3a ? 0 ,∴ ? x ?3?? x ? a? ? 0

(1)当 a ? 3时, x ? a或x ? 3 ? ,不等式解集为 x x ? a或x ? 3?;

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(2)当 a ? 3时,不等式为 ? x ? 3?2 ? 0 ,解集为?x x ? R且x ? 3? ;

(3)当 a ? 3时, x ? 3或x ? a ? ,不等式解集为 x x ? 3或x ? a?。

·················································································································10 分

18.解:(Ⅰ) f (x) ? 3 sin 2x ? cos2x ?1? m ? 2sin(2x ? ? ) ? m ?1 6



x

?

???0,?2

? ??



2x

?

? 6

?

?? ?? 6

,7? 6

? ??

,于是有

2

?

m

?1

?

6

,解得

m

?

3



··················································································································6 分

(Ⅱ) f (x) ? 2sin(2x ? ? ) ? 4(x ? R) 的最小值为 ? 2 ? 4 ? 2 ,此时相应的 x 的取值集合由 6

2x

?

? 6

?

3? 2

?

2k? (k

? ? Z ) 求得,为 ?x
?

x

?

2? 3

? k?,k

?

Z

? ?



?

·················································································································12 分

19.解:可求得 an

?

??2 ? ?2n

, (n ? 5,

? 1) (n ?

2,

n

?

N*

)

。令

an

?

0 ,即 2n ? 5 ?

0,得 n

?

5 2



··················································································································4 分

?当 n ? 2 时,Tn ? ?Sn ? ?n2 ? 4n ?1 ; 当n ? 2时

Tn ? a1 ? a2 ? a3 ??? an ? ?a1 ? a2 ? a3 ??? an ? ?2(a1 ? a2 ) ? Sn ? n2 ? 4n ? 7
·················································································································10 分

?? n2 ? 4n ?1,n ? 2

? Tn

?

? ?

n2

?

4n

? 7,n

?

2



·················································································································12 分

20.解:(Ⅰ)由余弦定理及已知条件得, a2 ? b2 ? ab ? 4 ,

又因为 △ABC 的面积等于 3 ,所以 1 ab sin C ? 3 ,得 ab ? 4 . 2

联立方程组

?a2 ? ?ab

? b2 ? ? 4,

ab

?

4,解得

a

?

2



b

?

2



··················································································································6 分

(Ⅱ)由题意得 sin(B ? A) ? sin(B ? A) ? 4sin Acos A ,

即 sin Bcos A ? 2sin Acos A ,

当 cos A ? 0 时, A ? ? , B ? ? , a ? 4 3 , b ? 2 3 ,

2

6

3

3

当 cos A ? 0 时,得 sin B ? 2sin A ,由正弦定理得 b ? 2a ,

联立方程组

?a ??b

2 ? b2 ? ? 2a,

ab

?

4,解得 a

?

23 3

,b

?

43 3



所以 △ABC 的面积 S ? 1 ab sin C ? 2 3 。

2

3

·················································································································12 分

21.解:设第 n 个月月底的余额为 an 元,则 a1 ?11260 , an?1 ? an ? (1? 20%) ? an ? 20%?10% ? 540 ? 1.18an ? 540 ,
··················································································································4 分

于是 a12 ? 1.18a11 ? 540 ? 1.18?1.18a10 ? 540? ? 540 ? 1.182 a10 ? ?1.18 ?1? ? 540 ?

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? ? =1.1811a1 ? 1.1810 ?1.189 ?

?1.18 ?1 ? 540 =1.1811 ?11260 ? 1.1811 ?1 ? 540 ? 54046.8 . 1.18 ?1

··················································································································8 分

还清银行贷款后剩余资金为 a12 ?10000 ? ?1? 5%? ? 54046.8 ?10500 ? 43546.8 .

答:到这年年底该个体户还贷款前尚余资金 54046.8 元;还清银行贷款后还有资金 43546.8元.

·················································································································12 分

22.解: (Ⅰ) 证明:由题意知: an?1 ? an2 ? 2an ∴ an?1 ?1 ? (an ?1)2

∵ a1 ? 9 ∴ an ?1 ? 0 ∴ lg(an?1 ?1) ? lg(an ?1)2 即 bn?1 ? 2bn ,

又∵ b1 ? lg(1? a1) ? 1 ? 0 ∴?bn? 是公比为 2 的等比数列。

··················································································································4 分

(Ⅱ) 由(1)知: bn ? b1 ? 2n?1 ? 2n?1

∴ cn ? n ? 2n?1 。

∴ Sn ? c1 ? c2 ? ? ? cn ? 1? 20 ? 2 ? 21 ? 3? 22 ? ?? n ? 2n?1

∴ 2Sn ? 1? 21 ? 2 ? 22 ? 3 ? 23 ? ? ? (n ? 1) ? 2n?1 ? n ? 2n

∴ ?Sn ? 1? 20 ? 21 ? 22 ?

? 2n?1 ? n ? 2n ? 1? 2n ? n ? 2n ? 2n ?1? n ? 2n 1? 2

∴ S n ? n ? 2n ? 2n ?1 。

··················································································································8 分

(Ⅲ) ∵ an?1 ? an2 ? 2an ? an (an ? 2) ? 0

∴ 1 ?1(1 ? 1 ) an?1 2 an an ? 2

∴ 1 ?1? 2 an ? 2 an an?1

∴ dn

?

1 an

?

1 an

?

2 an?1

?

2( 1 an

?

1) an?1

∴ Dn

?

d1

? d2

??? dn

?

1 2(
a1

?

1 a2

?

1 a2

?

1 a3

?? 1 an

?

1 )
an?1

?

1 2(
a1

?

1 )
an?1

又由(1)知: lg(1 ? an) ? 2n?1

∴ an ?1 ? 102n?1

∴ an?1 ? 102n ?1



Dn

?

2(1 9

?

1 102 n

)。 ?1

·················································································································12 分

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