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

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 , 2x?1 ? 0 C. ?x ?R , lg x ? 1 1 3. 2 lg 2 ? lg 的值为 25 A.1 B.2 4. 函数 f ( x) ? (1 ?
1 ) sin x 的图象大致为 x2

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

C.3

D.4

A.

B.

C.

D.

???? ??? ? ??? 1 ??? ? ? ??? ? 5.△ABC 中,若 AD ? 2DB , 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
3 ) ,则 ? 的值可以是 2

g ( x) 的图象,若 f ( x) 、 g ( x) 的图象都经过点 P(0,

? 5? 5? ? 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
A. 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 D.1055
?? x 2 ? 1, ?1 ≤ x ≤ 1 ? 10.定义在 R 上的函数 f ( x) 满足 f ( x ? 4) ? f ( x), f ( x) ? ? ,若关 ?log 2 (? | x ? 2 | ?2), 1 ? x ≤ 3 ?

于 x 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 1 1 1 A. ( , ) B. ( , ) C. (16 ? 6 7, ) 6 4 3 6 4

D. ( ,8 ? 2 15)

1 6

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 (3x ? 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若 f ( x) ? ?
x≥0 ?1, ,则 f ( x) ? A ; ??1, x ? 0

②若 f ( x) ? kx ,则 f ( x) ? A ; ③若 f ( x) ? A ,则 y ? f ( x) 可为奇函数; ④若 f ( x) ? A ,则对任意不等实数 x1 , x2 ,总有
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) 在△ ABC 中,角 A 、 B 、 C 的对边分别为 a 、 b 、 c ,设 S 为△ ABC 的面积,满足
4S ? 3(a 2 ? b2 ? c 2 ) .

(Ⅰ)求角 C 的大小; (Ⅱ)若 1 ?
??? ??? ? ? 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 ? ( 前 n 项和 Rn .
2 an ? ) ,令 cn ? an bn (n ? N ) .求数列 {cn } 的 2

19. (本小题满分 12 分)
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.

已知函数 f ( x) ? 4 x3 ? 3x2 sin ? ?

20. (本小题满分 13 分)

? ? 设平面向量 a ? ( 3 sin x,cos x ) , b ? (2sin( ? x),cos x) ,已知 f ( x) ? a ? b ? m 在 [0, ] 上 2 2 2 的最大值为 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) 已知函数 f ( x) ?
a 其中 a ? 0 F ? x ? (a ? 1) ln x ? 15a , ( x) ? ?2 x3 ? 3(a ? 2) x 2 ? 6 x ? 6a ? 4a 2 , x

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) 图象的对称中心坐标; (Ⅲ)设函数 g ( x) ? ?
?( F ( x) ? 6 x 2 ? 6(a ? 1) x) ? e x , x ≤ 1, ?e ? f ( x), x ? 1.

( e 是自然对数的底数) ,是否存在

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

泸州市 2014 届高三第一次教学教学质量诊断性考试
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. ?2 ; 三、解答题 16.解: (Ⅰ)样本的平均成绩 x ? 方差 s 2 ?
92 ? 98 ? 98 ? 85 ? 85 ? 74 ? 74 ? 74 ? 60 ? 60 ········· ·· ? 80 , ········· ··2 分 10

3 12. (0, ) ? (1, ??) ; 4

13. an ? 4n ?1 ;

14. (??, ?5] ;

15. ②③.

1 [(92 ? 80)2 ? (98 ? 80)2 ? (98 ? 80)2 ? (85 ? 80) 2 ?(85 ? 80)2 ?(74 ? 80)2 ? (74 ? 80)2 10

?(74 ? 80)2 ? (60 ? 80)2 ? (60 ? 80)2 ] ? 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 . ············································································ 分 ··········································································· 7
P(? ? 0) ?
2 2 C7 C1 C1 C3 7 7 1 ········· · ········ ? , P(? ? 1) ? 3 2 7 ? , P(? ? 2) ? 2 ? . ··········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 a2 ? b2 ? c2 ? 2ab cos C . ···································2 分 ································ ·· 2

因为 4S ? 3(a 2 ? b2 ? c 2 ) ,
1 所以 4 ? ab sin C ? 2 3ab cos C , ··········································3 分 ······································· ·· 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 , ············································7 分 ········································· ·· ? cos A sin B b sin C 2c 即 ·················································· ·· ? , ·················································· ··8 分 cos A sin B b

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: (Ⅰ)设等差数列 {an } 的公差为 d , ∵ a1 ? 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 ······························· ·· ) ? 1 ? ( )2 n ? 1 ? ( )n ,··································7 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 ? ( ) , ················································8 分 ············································· ·· 2 2n n 所以数列 {an } 的通项公式为 an ? 2n ;所以 cn ? n ? n ?1 , ····················9 分 ················· ·· 2 2 1 2 3 n 1 1 2 3 n 所以 Rn ? 0 ? 1 ? 2 ? ? ? n ?1 ,即 Rn ? ? 2 ? 3 ? ? ? n , ············· ·10 分 ············· · 2 2 2 2 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 x2 ? 6 x sin ? , ················································ ··1 分 ················································ ··

(Ⅱ)因为 Tn ? 1 ? (

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

当x?

所以 f ?( x) ? 12 x 2 ? 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 ····································· ·· ? 0 , f ( ) ? ? ? 0 , ····································· ··5 分 32 2 32 故函数 f ( x) 的零点个数有 3 个; ······································· ··6 分 ······································· ··

而 f (0) ?

(Ⅱ) f ?( x) ? 12x2 ? 6x sin? ,令 f ?( x) ? 0 ,得 x1 ? 0, x2 ?

sin ? , ·····················7 分 ·················· ·· 2 函数 f ( x) 存在极值, sin ? ? 0 , ······································ ··8 分 ······································ ··

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:

x

(??,0)

0

(0,

sin ? ) 2

sin ? 2

(

sin ? , ??) 2

f ?( x)

+ ↗

0 极大值

- ↘

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 ··············· · ) ? 0 ,必有 ? sin 3 ? ? ? 0 可得 0 ? sin ? ? ,··············· ·11 分 2 2 4 32

? 5? 所以 ? 的取值范围是 ? ? (0, ) ? ( , ? ) . ······························ ·12 分 ······························ · 6 6 ? 20.解: (Ⅰ) f ( x) ? 3 sin x ? 2sin( ? x) ? 2cos2 x ? m , ··································1 分 ······························· ·· 2 ············································· ·· ? 3 sin 2 x ? 2cos2 x ? 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] , ················································5 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:
5 ? ? 14 ? 3 ························· ·· 2sin[2( ? x0 ) ? ] ? 4 ? ,即 sin(2 x0 ? ) ? , ····························7 分 2 6 5 6 5 ? ? ? 2? 7? 因为 x0 ? [ , ] ,则 2 x0 ? ? [ , ] , ···································8 分 ································ ·· 6 3 6 4 2 6 2

?

? 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 x2 ? 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 x 2 ? 6(a ? 2) x ? 6 , ··········································5 分 ······································· ·· 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 F1 ( x) ? ?2 x3 ? 6 x 的图象关于 (0, 0) 对称, ························· ··7 分 ························· ·· 所以 F ( x) ? ?2 x3 ? 6 x ? 1 的图象的对称中心坐标为 (0, ?4) ; ····················8 分 ················· ·· (Ⅲ)假设存在 a 使 g ( x) 在 [a, ?a] 上为减函数, 设 h1 ( x) ? F ( x) ? 6 x2 ? 6(a ? 1) x) ? e x ? (?2 x3 ? 3ax2 ? 6ax ? 6a ? 4a 2 ) ? e x ,
a h2 ( x) ? e ? f ( x) ? e ? ( ? x ? (a ? 1) ln x ? 15a) , x ······························ ·· h1?( x) ? (?2 x3 ? 3(a ? 2) x 2 ? 12ax ? 4a 2 ) ? e x , ······························ ··9 分

设 m( x) ? (?2 x3 ? 3(a ? 2) x 2 ? 12ax ? 4a 2 ) , 当 g ( x) 在 [a, ?a] 上为减函数,则 h1 ( x) 在 [a,1] 上为减函数, h2 ( x) 在 ·································· · [1, ?a] 上为减函数,且 h1 (1) ≥ h2 (1) . ·································· ·10 分 由(Ⅰ)知当 a ? ?1 时, f ( x) 的单调减区间是 (1, ?a) , 由 h1 (1) ≥ h2 (1) 得: 4a2 ? 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) ? ?4a2 ? 12a ? 8 , 故 m(?2) ? ?4a 2 ? 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) ? ?4a2 ? 12a ? 8 ? 0 , 综上所述,符合条件的 a 满足 [?3, ?2] . ································14 分 ······························ ·


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