泸州市2014届高三第一次数学教学质量诊断性考试

泸州市 2014 届高三第一次教学教学质量诊断性考试 数学(理工类)
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 3. 2lg 2 ? lg A.1
1 的值为 25

B.{2,4} D.{2,4,8} B. ?x ? N* , ( x ? 1)2 ? 0 D. ?x ?R , tan x ? 2

B.2

C.3

D.4

1 4. 函数 f ( x) ? (1 ? 2 ) sin x 的图象大致为 x

A.

B.

C.

D.

???? ??? ? ??? 1 ??? ? ? ??? ? 5.△ABC 中,若 AD ? 2 DB , CD ? CA ? ? CB ,则 ? = 3 1 2 2 A. B. C. ? 3 3 3

D. ?

1 3

6. 将函数 f ( x) ? sin(2 x ? ? )(?

?
2

?? ?

?
2

) 的图象向右平移 ? (? ? 0) 个单位长度后得到函数 g ( x) 的图象,

3 ) ,则 ? 的值可以是 2 ? 5? 5? ? A. B. C. D. 2 3 6 6 7. 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

若 f ( x) 、 g ( x) 的图象都经过点 P(0,

A.充分而不必要条件 C.充分必要条件
? 1

B.必要而不充分条件 D.既不充分也不必要条件

8. 若曲线 f ( x ) ? x 2 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 18,则 a ? A.64 B.32 C.16 D.8 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 的方 ?log 2 (? | x ? 2 | ?2), 1 ? x ≤ 3 ?

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 C. (16 ? 6 7, ) 6 1 D. ( ,8 ? 2 15) 6

1

二、填空题:本大题共 4 小题,每小题 5 分,共 25 分. 11.复数 (m2 ? 5m ? 6) ? (m2 ? 2m ? 15)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 14.设 f(x)是定义在 R 上的奇函数,且当 x ≥ 0 时, f ( x) ? x 2 ,若对任意 x ? [ a, a ? 2],不等式
f ( x ? a) ≥ f (3 x? 1) 恒成立,则实数 a 的取值范围是
2 2



15.已知集合 A ? { f ( x) | f ( x) ? f ( y) ? f ( x ? y) ? f ( x ? y),x、y ? R} ,有下列命题: ① f ( x) ? ? 若
x≥0 ?1, ,则 f ( x) ? A ; ??1, x ? 0

② f ( x) ? kx ,则 f ( x) ? A ; 若 ③ f ( x) ? A ,则 y ? f ( x) 可为奇函数; 若 ④ f ( x) ? A ,则对任意不等实数 x1 , x2 ,总有 若
f ( x1 ) ? f ( x2 ) ? 0 成立. x1 ? x2

其中所有正确命题的序号是 . (填上所有正确命题的序号) 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. (本小题满分 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 10 个样本中随机抽出 2 名学生的成绩,设选出学生 的分数为 90 分以上的人数为 X,求随机变量 ? 的分布列和均值.

17. (本小题满分 12 分)
b c 设 在△ABC 中, A 、 、 的对边分别为 a 、 、 , S 为△ABC 的面积, 角 满足 4S ? 3(a2 ? b2 ? c2 ) . B C

(Ⅰ )求角 C 的大小; (Ⅱ 1 ? )若
??? ??? ? ? tan A 2c ,且 AB?BC ? ?8 ,求 c 的值. ? tan B b

18. (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 . (Ⅰ )求数列 {an } 的通项公式; (Ⅱ )设数列 {bn } 前 n 项和为 Tn ,且 Tn ? 1 ? (
Rn .

2 an ? ) ,令 cn ? anbn (n ? N ) .求数列 {cn } 的前 n 项和 2

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

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

20. (本小题满分 13 分)
2

? ? 设平面向量 a ? ( 3 sin x, x) , b ? (2sin( ? x),cos x) ,已知 f ( x) ? a ? b ? m 在 [0, ] 上的最大值 2cos 2 2 为 6. (Ⅰ )求实数 m 的值; ? ? ? 14 (Ⅱ )若 f ( ? x0 ) ? , x0 ? [ , ] .求 cos 2x0 的值. 4 2 2 5

21. (本小满分 14 分) a 已 知 函 数 f ( x) ? ? x ? ( a ? 1) l nx ? 1 5 F ( x) ? ?2 x3 ? 3(a ? 2) x2 ? 6 x ? 6a ? 4a 2 , 其 中 a ? 0 且 , a x 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) 图象的对称中心坐标; ) (Ⅲ 设函数 g ( x) ? ? )
?( F ( x) ? 6 x 2 ? 6(a ? 1) x) ? e x , x ≤ 1, ( e 是自然对数的底数) 是否存在 a 使 g ( x) , x ? 1. ?e ? f ( x),

在 [a, ?a] 上为减函数,若存在,求实数 a 的范围;若不存在,请说明理由.

3

一、选择题 题号 答案 二、填空题 11. ?2 ; 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 ······ · 2 ····· · ? 80 , ········ 分 10
3 12. (0, ) ? (1, ??) ; 4

1 D

2 B

3 B

4 A

5 B

6 B

7 C

8 A

9 C

1 0 D

13. an ? 4n ?1 ;

14. (??, ?5] ;

15. ② . ③

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 ] ··························· 分 ··········· ·········· ···· · 4 ·········· ··········· ··· ·

··········· ·········· ··········· ······ ·· ·········· ··········· ··········· ······ · ? 175 ; ········································ 6 分 (Ⅱ )由题意知选出学生的分数为 90 分以上的人数为 ? ,得到随机变量 X ? 0,1, 2 .7 分
P(? ? 0) ?
2 C7 C1 C1 C2 7 7 1 ? , P(? ? 1) ? 3 2 7 ? , P(? ? 2) ? 23 ? . ··········10 分 ········· · ········ 2 15 C10 15 C10 C10 15

?

0
7 15

1
7 15

2
1 15

P
E? ? 0 ?

7 7 1 3 ··········· ·········· ····· ·········· ··········· ····· ? 1? ? 2 ? ? . ··························· 12 分 15 15 15 5

1 17.解:(Ⅰ S ? ab sin C ,且 a 2 ? b2 ? c 2 ? 2ab cos C . ······················· 分 ) ··········· ·········· · 2 ·········· ·········· · 2

因为 4S ? 3(a2 ? b2 ? c2 ) ,
1 所以 4 ? ab sin C ? 2 3ab cos C ,························· ·· 分 ··········· ·········· ···· · 3 ·········· ··········· ···· · 2

所以 tan C ? 3 ,··································· 4 分 ··········· ·········· ··········· · ·· ·········· ··········· ··········· · · 因为 0 ? C ? ? , 所以 C ? (Ⅱ )由 1 ?
π ; ··········· ··········· ·········· ···· · 6 分 ··········· ·········· ··········· ··· ·· ·········· ··········· ··········· ··· · 3

tan A 2c 得: ? tan B b cos A sin B ? sin A cos B 2c , ··········· ··········· ···· ·· 分 ··········· ·········· ····· · 7 ·········· ··········· ····· · ? cos A sin B b sin C 2c 即 ··········· ·········· ··········· · 8 ·········· ··········· ·········· · ? , ··········· ··········· ·········· ·· 分 cos A sin B b

4

又由正弦定理得 cos A ?

1 , ··········· ··········· ······ · 9 分 ··········· ·········· ······ ·· ·········· ··········· ······ · 2

∴A ? 60? , ∴ ABC 是等边三角形, ····························· ·10 分 △ ··········· ·········· ········ · ·········· ··········· ········ ??? ??? ? ? ? ∴AB?BC ? c ? c ? cos120 ? ?8 , ···························11 分 ··········· ·········· ····· · ·········· ··········· ···· 所以 c ? 4 . ····································· ·12 分 ··········· ·········· ··········· ····· · ·········· ··········· ··········· ····· 18.解: )设等差数列 {an } 的公差为 d , (Ⅰ ∵a1 ? 2d ? 6 , 2a1 ? 9d ? 22 , ····························· 分 ··········· ·········· ······ · 2 ·········· ··········· ····· · ∴a1 ? 2 , d ? 2 , ·································· ·· 分 ··········· ·········· ··········· ·· · 4 ·········· ··········· ··········· ·· · ··········· ····· ·· ·········· ······ · 所以数列 {an } 的通项公式 an ? 2 ? ? n ? 1? ? 2 ? 2n ; ·················· 6 分 (Ⅱ )因为 Tn ? 1 ? (
2 an 2 1 ··········· ········· · 7 ·········· ·········· · ) ? 1 ? ( ) 2 n ? 1 ? ( ) n , ··········· ········· ·· 分 2 2 2 2 1 当 n ? 1 时, a1 ? T1 ? 1 ? ( )2 ? , 2 2 1 n 1 1 当 n ≥ 2 时, an ? Tn ? Tn ?1 ? 1 ? ( ) ? 1 ? ( )n ?1 ? ( )n , 2 2 2 1 n 且 n ? 1 时满足 an ? ( ) , ······························· 分 ··········· ·········· ········ · 8 ·········· ··········· ······· · 2 所以数列 {an } 的通项公式为 an ? 2n ;

2n n ··········· ·········· ·········· ·· ·········· ··········· ·········· · ? n ?1 , ································· 9 分 n 2 2 1 2 3 n 所以 Rn ? 0 ? 1 ? 2 ? ? ? n ?1 , 2 2 2 2 1 1 2 3 n 即 Rn ? ? 2 ? 3 ? ? ? n , ·························· ·10 分 ··········· ·········· ····· · ·········· ··········· ····· 2 2 2 2 2 1 1? n 1 1 1 1 1 n 2 ? n ? 2 ? n ? 2 , ···· ·11 分 两式相减得: Rn ? 0 ? 1 ? 2 ? ? ? n ?1 ? n ? ···· · ···· 1 2n 2 2 2 2 2 2 2n 1? 2 n?2 ··········· ·········· ··········· · ·········· ··········· ·········· 所以 Rn ? 4 ? n ?1 . ·································12 分 2 19.解: (I) f ?( x) ? 12 x 2 ? 6 x sin ? , ································ 1 分 ··········· ·········· ········· ·· ·········· ··········· ········· ·

所以 cn ?

sin ? 3 时, f ?( x) 有最小值为 f ?( x) ? ? sin 2 ? , 4 4 3 3 所以 ? sin 2 ? ? ? ,即 sin 2 ? ? 1 , ························· 2 分 ··········· ·········· ·· ·· ·········· ··········· ·· · 4 4 因为 ? ? [0, ? ] ,所以 sin ? ? 1 ,·························· ·· 分 ··········· ·········· ····· · 3 ·········· ··········· ····· ·

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, ······· ·· 分 ······· · 4 ······· · 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; ··························· 分 ··········· ·········· ···· · 6 ·········· ··········· ··· · sin ? (Ⅱ f ?( x) ? 12 x 2 ? 6 x sin ? ,令 f ?( x) ? 0 ,得 x1 ? 0, x2 ? ) , ··········· · ·· 分 ··········· · · 7 ·········· · · 2 函数 f ( x) 存在极值, sin ? ? 0 , ························ ·· 分 ··········· ·········· ··· · 8 ·········· ··········· ··· · 由 ? ? [0, ? ] 及(I) ,只需考虑 sin ? ? 0 的情况. 当 x 变化时, f ?( x) 的符号及 f ( x) 的变化情况如下表:

5

x
f ?( x)

(??, 0)

0 0 极大值

(0,

sin ? ) 2

sin ? 2

(

sin ? , ??) 2

+ ↗

- ↘

0 极小值

+ ↗

f ( x)

因此,函数 f ( x) 在 x ? 要使 f (

sin ? sin ? 1 1 处取得极小值 f ( ····· · ···· ) ? ? sin3 ? ? , ······10 分 2 2 4 32

sin ? 1 1 1 ········· ·11 ········· ) ? 0 ,必有 ? sin3 ? ? ? 0 可得 0 ? sin ? ? ,··········· 分 2 2 4 32

? 5? 所以 ? 的取值范围是 ? ? (0, ) ? ( , ? ) . ··················· ·12 分 ··········· ········ · ·········· ········· 6 6 ? 20.解: ) f ( x) ? 3 sin x ? 2sin( ? x) ? 2cos 2 x ? m , ······················ 1 分 (Ⅰ ··········· ········· ·· ·········· ·········· · 2 ? 3 sin 2 x ? 2cos 2 x ? m , ······························ 2 分 ··········· ·········· ······· ·· ·········· ··········· ······· · ? ··········· ·········· ··········· · 3 ·········· ··········· ·········· · ? 2sin(2 x ? ) ? 1 ? m ·································· 分 6 ? ? ? 7? ··········· ·········· ····· ·· ·········· ··········· ····· · ? x ? [0, ], 2 x ? ? [ , ] , ···························· 4 分 2 6 6 6 ? 1 ··········· ·········· ········ · 5 ·········· ··········· ······· · ? 2sin(2 x ? ) ?[? ,1] , ······························· 分 6 2 ? f ( x)max ? 2 ? 1 ? m ? 6,? m ? 3 , 所以 m ? 3 ; ····································· 6 分 ··········· ·········· ··········· ··· ·· ·········· ··········· ··········· ··· · ? ? 14 (Ⅱ )因为 f ( x) ? 2sin(2 x ? ) ? 4 ,由 f ( ? x0 ) ? 得: 6 2 5 ? ? 14 ? 3 ··········· ····· · 7 ·········· ······ · 2sin[2( ? x0 ) ? ] ? 4 ? ,即 sin(2 x0 ? ) ? , ················ ·· 分 2 6 5 6 5 ? ? ? 2? 7? 因为 x0 ? [ , ] ,则 2 x0 ? ?[ , ] ,······················· 分 ··········· ·········· · 8 ·········· ·········· · 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 ?? ? ? ? ? . ··························· ·12 分 ··········· ·········· ······ · ·········· ··········· ······ 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 x ? 1 时,函数 F ( x) 有极值, )当 所以 F ?( x) ? ?6 x2 ? 6(a ? 2) x ? 6 ,························· ·· 分 ··········· ·········· ···· · 5 ·········· ··········· ···· ·
6

且 F ?(1) ? 0 ,即 a ? ?2 , ······························· 6 分 ··········· ·········· ········ ·· ·········· ··········· ········ · 所以 F ( x) ? ?2 x3 ? 6 x ? 4 ,
F ( x) ? ?2 x3 ? 6 x ? 1 的图象可由 F1 ( x) ? ?2 x3 ? 6 x 的图象向下平移 4 个单位长度得到,而

··········· ·········· · 7 ·········· ·········· · F1 ( x) ? ?2 x3 ? 6 x 的图象关于 (0, 0) 对称, ······················· 分 所以 F ( x) ? ?2 x3 ? 6 x ? 1 的图象的对称中心坐标为 (0, ?4) ; ··········· ·· 分 ··········· · 8 ·········· · · (Ⅲ )假设存在 a 使 g ( x) 在 [a, ?a] 上为减函数, 设 h1 ( x) ? F ( x) ? 6 x 2 ? 6(a ? 1) x) ? e x ? (?2 x3 ? 3ax 2 ? 6ax ? 6a ? 4a 2 ) ? e x , a h2 ( x) ? e ? f ( x) ? e ? ( ? x ? (a ? 1)ln x ? 15a) , x 3 ··········· ········ · 9 ·········· ········· · h1?( x) ? (?2 x ? 3(a ? 2) x2 ? 12ax ? 4a 2 ) ? e x , ··················· ·· 分 设 m( x) ? (?2 x3 ? 3(a ? 2) x2 ? 12ax ? 4a 2 ) , h 当 g ( x) 在 [a, ?a] 上为减函数, h1 ( x) 在 [a,1] 上为减函数, 2 ( x) 在 [1, ? a ] 上为减函数, 则 且 h1 (1) ≥ h2 (1) . ·································· 分 ································ ·10 ·········· ··········· ··········· 由(Ⅰ )知当 a ? ?1时, f ( x) 的单调减区间是 (1, ? a ) , 由 h1 (1) ≥ h2 (1) 得: 4a 2 ? 13a ? 3 ≤ 0 ,
1 解得: ?3 ≤ a ≤ ? , ····························· ·11 分 ··········· ·········· ········ · ·········· ··········· ········ 4 当 h1 ( x) 在 [a,1] 上为减函数时,对于 ?x ? [a,1] , h1?( x) ≤ 0 即 m( x) ≤ 0 恒成立, 因为 m?( x) ? ?6( x ? 2)( x ? a) , (1)当 a ? ?2 时, m( x) 在 [ a , ?2] 上是增函数,在 (??, a],[?2, ??) 是减函数,

所以 m( x) 在 [a,1] 上最大值为 m(?2) ? ?4a 2 ? 12a ? 8 , 故 m(?2) ? ?4a2 ? 12a ? 8 ≤ 0 , 即 a ≤ ?2 ,或 a ≥ ?1 ,故 a ? ?2 ; ····················· ·12 分 ··········· ·········· · ·········· ··········· (2)当 a ? ?2 时, m( x) 在 [ ?2, a ] 上是增函数,在 (??, ?2],[a, ??) 是减函数, 所以 m( x) 在 [a,1] 上最大值为 m(a) ? a 2 (a ? 2) , 故 m(a) ? a 2 (a ? 2) ≤ 0 ,则 a ≤ ?2 与题设矛盾; ·············· ·13 分 ··········· ··· · ·········· ···· (3)当 a ? ?2 时, m( x) 在 [ ?2,1] 上是减函数, 所以 m( x) 在 [a,1] 上最大值为 m(?2) ? ?4a 2 ? 12a ? 8 ? 0 , 综上所述,符合条件的 a 满足 [?3, ?2] . ····················· 分 ··················· ·14 ·········· ·········

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