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四川省泸州市2014届高三第一次教学质量诊断性考试数学文试题 Word版含答案


2014 届高三第一次教学教学质量诊断性考试 数学(文史类)
一、选择题:本大题共有 10 个小题,每小题 5 分,共 50 分.每小题给出的四个选项中,只 有一项是符合要求的. 1.已知全集 U={1,2,3,4,5,6,7,8},M ={1,3,5,7},N ={5,6,7},则 ? (M ? N ) = U A.{5,7} C.{1,3,5,6,7} 2. 下列命题中的假命题是 A. ?x ?R , 2 x ?1 ? 0 C. ?x ?R , lg x ? 1 1 3. 2lg 2 ? lg 的值为 25 A.1 B.2 4.下列函数与 y ? x 相等的是 A. y ? ( 3 x )3 C. y ? ( x )2 B.{2,4} D.{2,4,8} B. ?x ? N ? , ( x ? 1)2 ? 0 D. ?x ?R , tan x ? 2

C.3 B. y ?
x2 x

D.4

D. y ? x 2

???? ??? ? ??? 1 ??? ? ? ??? ? 5.△ABC 中,若 AD ? 2 DB , CD ? CA ? ? CB ,则 ? = 3 1 2 2 1 A. B. C. ? D. ? 3 3 3 3 6.若曲线 f ( x) ? x 2 ( x ? 0) 在点 (a, f (a)) 处的切线与两条坐标轴围成的三角形的面积为 54, 则a? A.3 B.6 C.9 D.18 ? 7.如图为函数 f ( x) ? A sin(? x ? ? ) (其中 ? ? 0,0 ≤ ? ≤ )的部分图象,其中 A,B 两点之 2 间的距离为 5 ,那么 f (?1)=

y 2 1 O -2 x A

3 2 C. ?1

A. ?

1 2 D. 1

B. ?

8.设数列 {an } 是首项大于零的等比数列,则“ a1 ? a2 ”是“数列 {an } 是递增数列”的 A.充分而不必要条件 B.必要而不充分条件 C.充分必要条件 D.既不充分也不必要条件

B

9.一支人数是 5 的倍数且不少于 1000 人的游行队伍,若按每横排 4 人编队,最后差 3 人; 若按每横排 3 人编队,最后差 2 人;若按每横排 2 人编队,最后差 1 人.则这只游行队 伍的最少人数是 A.1025 B.1035 C.1045
2

D.1055

10.定义在 R 上的函数 f ( x) 满足 f ( x ? 4) ? f (x ), f (x )? ?

?? x ? 1, ? 1≤ x ≤ 1 ,若关于 x ?? | x ? 2 | ? 1, 1? x ≤ 3.

的方程 f ( x) ? ax ? 0 有 5 个不同实根,则正实数 a 的取值范围是
1 1 A. ( , ) 4 3 1 1 B. ( , ) 6 4

1 1 C. (16 ? 6 7, ) D. ( ,8 ? 2 15) 6 6 二、填空题:本大题共 4 小题,每小题 5 分,共 25 分. 11.复数 (m2 ? 3m ? 2) ? (m2 ? 4)i ( i 是虚数单位)是纯虚数,则实数 m 的值为 12. 等比数列 {an } 中, 若公比 q ? 4 , 且前 3 项之和等于 21, 则该数列的通项公式 an ?

. .

13.使不等式 log a

3 ? 1 (其中 0 ? a ? 1 )成立的 a 的取值范围是 4


1 x ? 1 ,则不等 2

14. 设函数 f ( x) 是定义在 (??, 0) ? (0, ??) 上的奇函数,且当 x ? 0 时, f ( x) ? 式 f ( x) ? x 的解集用区间表示为_________.

15 . 定 义 : 如 果 函 数 y ? f ( x) 在 定 义 域 内 给 定 区 间 [a ,b ]上 存 在 x0 (a ? x0 ? b) , 满 足
f ( x0 ) ? f (b )? f ( ) a , 则称函数 y ? f ( x) 是 [a, b] 上的“平均值函数”,x 0 是它的一个均值点, b?a

如 y ? x 4 是 [ ?1,1] 上的平均值函数,0 就是它的均值点.现有函数 f ( x) ? ? x2 ? mx ? 1 是
[ ?1,1] 上的平均值函数,则实数 m 的取值范围是



三、解答题:本大题共 6 小题,共 75 分. 解答应写出文字说明、证明过程或演算步骤. 16. (本小满分 12 分) 在一次数学统考后,某班随机抽取 10 名同学的成绩进行样本分析,获得成绩数据的茎 叶图如下. (Ⅰ )计算样本的平均成绩及方差; (Ⅱ )现从 80 分以上的样本中随机抽出 2 名学生,求抽出的 2 名学生的成绩分别在 [80,90) 、 [90,100] 上的概率.

17. (本小满分 12 分) (本小题满分 12 分) 设等差数列 {an } 的前 n 项和为 S n ,且 a3 ? 6 , S10 ? 110 .设数列 {bn } 前 n 项和为 Tn ,且
Tn ? 1 ? ( 2 an ) ,求数列 {an } 、 {bn } 的前 n 项和公式. 2

18. 在△ ABC 中,角 A 、 B 、 C 的对边分别为 a 、 b 、 c ,满足 a 2 ? b 2 ? c 2 ? ab ? 0 . (Ⅰ )求角 C 的大小; (Ⅱ )若
??? ??? ? ? sin C 2c ABC 的面积. ? ,且 AB?BC ? ?8 ,求△ cos A sin B b

19. (本小满分 12 分) 已知函数 f ( x) ? 4 x3 ? 3x2 sin ? ?
1 ,其中 x ?R , ? ? (0, ? ) . 32

3 (Ⅰ f ?( x) 的最小值为 ? ,试判断函数 f ( x) 的零点个数,并说明理由; )若 4 (Ⅱ )若函数 f ( x) 的极小值大于零,求 ? 的取值范围.

20. (本小满分 12 分) 设平面向量 a ? ( 3 sin( ? x ), 2 cos , b ? (?2cos x,cos x) ,已知函数 f ( x) ? a ? b ? m 在 ? x ) ? [0, ] 上的最大值为 6. 2 (Ⅰ )求实数 m 的值; ? ? 26 (Ⅱ )若 f ( x0 ) ? , x0 ? [ , ] .求 cos 2x0 的值. 4 2 5

21. (本小满分 14 分) a 已知函数 f ( x) ? ? x ? (a ? 1)ln x ? 15a , F ( x) ? 2 x3 ? 3(2a ? 3) x2 ? 12(a ? 1) x ? 12a ? 2 ,其 x 中 a ? 0 且 a ? ?1. (Ⅰ 当 a ? ?2 ,求函数 f ( x) 的单调递增区间; ) (Ⅱ 若 x ? ?1时,函数 F ( x) 有极值,求函数 F ( x) 图象的对称中心的坐标; ) (Ⅲ )设函数 g ( x) ? ?
? F ( x), ? f ( x), x ≤1, ( e 是自然对数的底数) ,是否存在 a 使 g ( x) 在 [a, ?a] 上 x ? 1.

为减函数,若存在,求实数 a 的范围;若不存在,请说明理由.

一、选择题

题 号 答 案

1 D

2 B

3 B

4 A

5 B

6 B

7 C

8 A

9 C

1 0 D

二、填空题 11.1; 三、解答题 16.解: )样本的平均成绩 x ? (Ⅰ 方差 s 2 ?
92 ? 98 ? 98 ? 85 ? 85 ? 74 ? 74 ? 74 ? 60 ? 60 ·········· ? 80 , ·········· 2 分 10
3 12. an ? 4n ?1 ; 13. (0, ) ; 4

14. (??, ?2) ? (0, 2) ;

15. (0, 2) .

1 [(92 ? 80)2 ? (98 ? 80)2 ? (98 ? 80)2 ? (85 ? 80)2 10

?(85 ? 80)2 ?(74 ? 80)2 ? (74 ? 80)2 ?(74 ? 80)2 ? (60 ? 80)2 ? (60 ? 80)2 ] ······································· 4 分 ·······································

···························································· ? 175 ; ·····························································6 分 (Ⅱ )从 80 分以上的样本中随机抽出 2 名学生,共有 10 种不同的抽取方法, ····· 8 分 ····· [90,100] 上共有 6 中不同的抽取方法, 而抽出的 2 名学生的分数分别在 [80,90) , 因此所求的概率为
6 3 ··········································· 12 ? . ············································ 分 10 5

17.解:设等差数列 {an } 的公差为 d , a ∵ 1 ? 2d ? 6 , 2a1 ? 9d ? 22 ,·············································· 2 分 ·············································· a1 ? 2 , d ? 2 , ··························································4 分 ∴ ························································· {an } 的通项公式 an ? 2 ? ? n ? 1? ? 2 ? 2n ; ······························6 分 ····························· 所以数列
2 an 2 1 ·································· ) ? 2 ? ( )2n ? 2 ? ( )n , ···································7 分 2 2 2 2 3 当 n ? 1 时, b1 ? T1 ? 2 ? ( )2 ? ,········································· 8 分 ········································· 2 2 1 1 1 当 n ≥ 2 时, bn ? Tn ? Tn ?1 ? 2 ? ( )n ? 2 ? ( )n ?1 ? ( )n , ························· 10 分 ························· 2 2 2 1 且 n ? 1 时不满足 bn ? ( )n , ················································ 分 ···············································11 2 ?3 n ?1 ? ? 所以数列 {bn } 的通项公式为 bn ? ? 2 .································· 分 ································12 ?( 1 ) n n ≥ 2 ? 2 ?

因为 Tn ? 2 ? (

18.解:(Ⅰ (Ⅰ ) )因为 a 2 ? b 2 ? c 2 ? ab ? 0 , 所以 a 2 ? b 2 ? c 2 ? ab , ················································ 1 分 ················································

所以 cos C ?

a 2 ? b2 ? c 2 ab 1 ····································· ? ? ,······································3 分 2ab 2ab 2

因为 0 ? C ? ? , ····················································· 5 分 ····················································· 所以 C ? (Ⅱ )由
π ; ························································ 6 分 ························································ 3

sin C 2c ? 正弦定理得: cos A sin B b c 2c ······················································· ? , ······················································· 7 分 b cos A b 1 ························································· cos A ? , ··························································8 分 2 A ∴ ? 60? , ∴ABC 是等边三角形, ················································ 分 △ ···············································10 ??? ??? ? ? ? AB ∴ ?BC ? c ? c ? cos120 ? ?8 , ∴ ? 4 , ···························································· 分 ··························································· 11 c 1 ABC 的面积 S ? c2 sin 60? ? 4 3 . ··································· 分 ·································· 12 所以△ 2 19.解: (I) f ?( x) ? 12 x 2 ? 6 x sin ? , ················································ 1 分 ················································
sin ? 3 时, f ?( x) 有最小值为 f ?( x) ? ? sin 2 ? , 4 4 3 2 3 2 所以 ? sin ? ? ? ,即 sin ? ? 1 , ····································· 2 分 ····································· 4 4 因为 ? ? (0, ? ) ,所以 sin ? ? 1 ,········································· 3 分 ·········································

当x?

所以 f ?( x) ? 12 x2 ? 6 x , 1 1 所以 f ( x) 在 (0, ) 上是减函数,在 (??,0),( , ??) 上是增函数, ··············4 分 ············· 2 2 1 1 7 而 f (0) ? ····································· ? 0 , f ( ) ? ? ? 0 , ····································· 5 分 32 2 32 故函数 f ( x) 的零点个数有 3 个; ······································· 6 分 ······································· (Ⅱ )
sin ? , ·····················7 分 ···················· 2 由 ? ? (0, ? ) 知 sin ? ? 0 ,根据(I) ,当 x 变化时, f ?( x) 的符号及 f ( x) 的变化情况如下表:
f ?( x) ? 12 x 2 ? 6 x sin ? 令 f ?( x) ? 0 ,得 x1 ? 0, x2 ?
(??, 0)

x
f ?( x)

0 0 极大值

(0,

sin ? ) 2

sin ? 2

(

sin ? , ??) 2

+ ↗

- ↘

0 极小值

+ ↗

f ( x)

sin ? sin ? 1 1 处取得极小值 f ( ········· ) ? ? sin3 ? ? , ··········9 分 2 2 4 32 sin ? 1 1 1 要使 f ( ················10 ) ? 0 ,必有 ? sin3 ? ? ? 0 可得 0 ? sin ? ? , ················· 分 2 2 4 32

因此,函数 f ( x) 在 x ?

? 5? 所以 ? 的取值范围是 ? ? (0, ) ? ( , ? ) . ································12 分 ································ 6 6 20.解: ) f ( x) ? 3 sin(? ? x) ? (?2cos x) ? 2cos2 x ? m , (Ⅰ
? 3 sin 2 x ? cos 2 x ? 1 ? m ,·············································2 分 ············································

················································ ? 2sin(2 x ? ) ? 1 ? m , ················································ 3 分 6 ? ? ? 7? ∵ x ?[0, ], 2 x ? ?[ , ] , ··········································4 分 ········································· 2 6 6 6 ? 1 ∴ 2sin(2 x ? ) ?[? ,1] 6 2 ∴ f ( x)max ? 2 ? 1 ? m ? 6 , ············································· 5 分 ············································· ∴ m ? 3 ;·························································· 6 分 ·························································· ? (Ⅱ )因为 f ( x) ? 2sin(2 x ? ) ? 4 , 6 26 ? 26 ? 3 由 f ? x0 ? ? 得: 2sin(2 x0 ? ) ? 4 ? ,则 sin(2 x0 ? ) ? ,···············7 分 ·············· 5 6 5 6 5 ? ? ? 2? 7? 因为 x0 ? [ , ] ,则 2 x0 ? ?[ , ] , ·································8 分 ································ 4 2 6 3 6 ? 因此 cos(2 x0 ? ) ? 0 , 6 ? 4 所以 cos(2 x0 ? ) ? ? , ···············································9 分 ·············································· 6 5 ? ? 于是 cos 2 x0 ? cos[(2 x0 ? ) ? ] , ······································· 10 分 ······································· 6 6 ? ? ? ? ? cos(2 x0 ? )cos ? sin(2 x0 ? )sin 6 6 6 6
4 3 3 1 3?4 3 ?? ? ? ? ? . ··········································· 分 ·········································· 12 5 2 5 2 10

?

21.解:(Ⅰ (Ⅰ 当 a ? ?2 , ) ) 2 3 x 2 ? 3x ? 2 , ········································· 1 分 ········································· f ?( x) ? 2 ? 1 ? ? x x x2 设 f ?( x) ? 0 ,即 x 2 ? 3x ? 2 ? 0 , 所以 x ? 1 ,或 x ? 2 , ·················································2 分 ················································ f ( x) 单调增区间是 (0,1) , (2, ??) ;····································· 4 分 ····································· (Ⅱ 当 x ? ?1时,函数 F ( x) 有极值, ) 所以 F ?( x) ? 6 x2 ? 6(2a ? 3) x ? 12(a ? 1) , ·································· 5 分 ·································· 3 且 F ?(?1) ? 0 ,即 a ? ? , ·············································6 分 ············································ 2 所以 F ( x) ? 2 x3 ? 6 x ? 16 ,
F ( x) ? 2 x3 ? 6 x ? 16 的图象可由 F1 ( x) ? 2 x3 ? 6 x 的图象向下平移 16 个单位长度

得到,而 F1 ( x) ? 2 x3 ? 6 x 的图象关于(0,0)对称, ·······················7 分 ······················ 所以函数 F ( x) ? 2 x 3 ? 6 x ?16 的图象的对称中心坐标为 (0, ?16) ; ·············8 分 ············ g ( x) 在 [a, ?a] 上为减函数, (Ⅲ )假设存在 a 使
F ?( x) ? 6 x 2 ? 6(2a ? 3) x ? 12(a ? 1) ,
? 6( x ? 1)( x ? 2a ? 2) , ················································· 9 分 ·················································

当 g ( x) 在 [a, ?a] 上为减函数,则 F ( x) 在 [a,1] 上为减函数, f ( x) 在 [1, ? a ] 上为 减函数,且 F (1) ≥ f (1) ,则 a ≥ ?3 . ···································· 分 ··································· 10 由(Ⅰ )知当 a ? ?1时, f ( x) 的单调减区间是 (1, ? a ) , 1 (1)当 a ? ? 时, F ?( x) ? 6( x ? 1)2 ≥ 0 , F ( x) 在定义域上为增函数, 2 不合题意;·························································· 分 ························································· 11

1 (2)当 a ? ? 时,由 F ?( x) ? 0 得:1 ? x ? 2a ? 2 , F ( x) 在 (??,1] 上为增函数,则 2 在 [a,1] 上也为增函数,也不合题意;···································· 分 ··································· 12 1 (3)当 a ? ? 时,由 F ?( x) ? 0 得: 2a ? 2 ? x ? 1 , F ( x) 在 [2a ? 2,1] 上为减函数, 2 g ( x) 在 [a, ?a] 上为减函数,则 F ( x) 在 [a,1] 上为减函数,则: 如果 ············································ 13 2a ? 2 ? a ,所以 a ≤ ?2 . ············································· 分 综上所述,符合条件的 a 满足 [?3, ?2] . ································· 14 分 ·································


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