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2014年大连市高三双基测试 数学理试题_图文

2014 年大连市高三双基考试 数学(理科)参考答案及评分标准
说明: 一、本解答给出了一种或几种解法供参考,如果考生的解法与本解答不同,可根据试题的主要考查内 容比照评分标准制订相应的评分细则. 二、对解答题,当考生的解答在某一步出现错误时,如果后继部分的解答未改变该题的内容和难度, 可视影响的程度决定后继部分的给分,但不得超过该部分正确解答应得分数的一半;如果后继部分的解答 有较严重的错误,就不再给分. 三、解答右端所注分数,表示考生正确做到这一步应得的累加分数. 四、只给整数分数,选择题和填空题不给中间分.

一.选择题 1.B 2.A 二.填空题 13.

3.C

4.D

5.A

6.B

7.D

8.B

9.A

10.A

11.C

12.D

1 2

14. 80

15. 13?

16. ( ?8, ?7)

三.解答题 17.解:(Ⅰ) f ( x) ? cos x(sin x ? 3 cos x) ? sin x cos x ? 3 cos x
2

?

sin 2 x 3 cos 2 x 3 ? ? 2 2 2

? sin( 2 x ?
当 2x ?

?
3

)?

3 .···················································································································· 4 分 2

?
3

? 2k? ?

?
2

( k ? Z) ,

即 x ? {x | x ? k? ?

5? 3 .················································ 6 分 , k ? Z } 时, f ( x) 取最大值 1 ? 12 2

(Ⅱ) f ( ) ? ?
2

A 2

3 ? ? ,可得 sin( A ? ) ? 0 ,因为 A 为△ ABC 内角,所以 A ? .········· 8 分 2 3 3
2 2 2 2

由余弦定理 a ? b ? c ? 2bc cos A ? b ? c ? bc , 由 a ? 3, b ? c ? 2 3 ,解得 bc ? 1 .························································································ 10 分 所以 S ?ABC ?

1 3 .································································································ 12 分 bc sin A ? 2 4

18. (Ⅰ) 2 ? 2 列联表如下 甲 厂 优质品 非优质品 合计 400 100 500 乙 厂 300 200 500 合计 700 300 1000 ········································· 4 分

?2 ?

1000(400 ? 200 ? 100 ? 300) 2 ? 47.619 ? 10.828 500 ? 500 ? 700 ? 300

有 99.9%的把握认为“生产的零件是否为优质品与在不同分厂生产有关”. ······················ 6 分 (Ⅱ)分层抽样从乙厂抽取优质品 3 件,非优质品 2 件.

X 取值为 0,1,2 . P ( x ? 0) ?
1 1 2 C2 C3 3 C 32 C2 1 3 ? , P ( x ? 1 ) ? ? , P ( x ? 2 ) ? ? , 2 2 2 5 C 5 10 C5 C 5 10

所以 X 的分布列为

X

0

1

2

P

1 10

3 5

3 10

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

3 3 6 所以 EX ? 1 ? ? 2 ? ? . 5 10 5 (或者直接利用超几何分布的期望公式求得) ································································ 12 分

19. 解:(Ⅰ)取 BC 中点 O ,因为三角形 ABC 是等边三角形,所以 AO ? BC , 又因为面 BCC ' B ' ? 底面 ABC , AO ? 面 ABC ,面 BCC ' B '? 面 ABC = BC ,

所以 AO ? 面 BCC ' B ' ,又 BB' ? 面 BCC ' B ' , 所以 AO ? BB' .又 BB ' ? AC , AO ? AC ? A , AO ? 面 ABC , AC ? 面 ABC , 所以 BB ' ? 底面 ABC . ············································································································ 4 分

(Ⅱ) 取 B ' C ' 中 点 O ' , 所 以 OO' ? 底 面 ABC . 分 别 以

OC , OA, OO' 为 x, y, z 轴建立空间直角坐标系如图所示.
所 以 B ( ?1,0,0), E (0, 3 ,2), F (1,0,1) , 在 A' C ' 上 找 一 点

M (a, 3 (1 ? a ),3)
所以 BM ? ( a ? 1, 3 (1 ? a ),3), BE ? (1, 3 ,2), BF ? ( 2,0,1) , 设 一个法向量 n ? ( x, y, z ) . 面 BEF 的

?

? ? ? ? BE ? n ? 0 ? x ? 3 y ? 2 z ? 0 ?? 则? ,不妨令 x ? 1 ,则 n ? (1, 3 ,?2) . ? ? ? BF ? n ? 0 ? 2 x ? z ? 0

········· 8 分

BM 和 面 BEF 所 成 角 的 余 弦 值 为 ? | cos ? n , BM ?|? | 2a ? 2 | 8 4a ? 4a ? 13
2

58 , 则 8

6 . 8 ? 6 1 23 ,解得 a ? 或 a ? ? (舍). 8 2 2

所以

所以 A' C ' 的中点符合题意. ······································································································ 12 分

(I)设点 20.解: 得 即 因为点 M,N 在椭圆 所以 故

,则由 , ·····································································································2 分 上, ······················································································· 4 分

由题意知, 所以

, ,···················································································································· 6 分

(II)将曲线 C 与直线 l 联立: 消 y 得: 直线 l 与曲线 C 交于 A、B 两点,设



?

····························································································································· 7 分 ································································································· 8 分

点 O 到直线 AB:

的距离

,

AB ? (1 ? k 2 ) x3 ? x4 ? (1 ? k 2 )[( x3 ? x4 ) 2 ? 4 x3 x4 ]

··························································································· 10 分 . 当且仅当 m ? 30 ? m ,即 m ? 15 时取等号.
2 2 2

所以,三角形 OAB 面积的最大值为

. ················································································ 12 分

x ? ln( x ? 1) 21.解:(Ⅰ) f ' ( x) ? x ? 1 2 ,·············································································· 1 分 x x 设 g ( x) ? ? ln( x ? 1) ,不妨令 x ? ?1 , x ?1
则 g ' ( x) ?

1 1 ?x ,当 x ? ( ?1,0) 时, g ' ( x) ? 0 , g ( x) 为增函数;当 x ? (0,??) 时, ? ? 2 x ? 1 ( x ? 1) 2 ( x ? 1)

g ' ( x) ? 0 , g ( x) 为减函数.
所以 g ( x) ? g (0) ? 0 ,即 f '( x) ? 0 ,所以在 x ? ( ?1, 0) ? (0, ??) 时, f '( x) ? 0 所以 f ( x) 在区间 ( ?1,0), (0,??) 上为减函数.··········································································· 5 分

1 1 x ? 1 等价于 ln( x ? 1) ? kx 3 ? x 2 ? x ? 0 , 2 2 1 3 2 设函数 h( x) ? ln( x ? 1) ? kx ? x ? x ,对于函数 h( x) ,不妨令 x ? 0 . 2
(Ⅱ) f ( x) ? kx ?
2

所以 h(0) ? 0 ,

h' ( x ) ?

1 ? 3kx 3 ? x 2 ? 3kx 2 x 2 (?3kx ? 1 ? 3k ) ··························· 8 分 ? 3kx 2 ? x ? 1 ? ? x ?1 x ?1 x ?1

当 k ? 0 时,在 x ? [0,??) 时, h' ( x) ? 0 ,所以 h( x) 在 x ? [0,??) 为增函数,所以 h( x) ? h(0) ? 0 ,不 符合题意; 当 0?k ?

1 1 ? 3k 1 ? 3k , 在 x ? [0, ] 时 , h' ( x) ? 0 , 所 以 h( x) 在 x ? [0, ] 为增函数,所以 3 3k 3k

h( x) ? h(0) ? 0 ,不符合题意; 1 时,在 x ? [0,??) 时, h' ( x) ? 0 ,所以 h( x) 在 x ? [0,??) 为减函数,所以 h( x) ? h(0) ? 0 , 即 3 1 ln( x ? 1) ? kx 3 ? x 2 ? x ? 0 在 x ? 0 上成立,符合题意; 2 1 综上,实数 k 的最小值为 .····································································································· 12 分 3
当k ? C 22.证明: (Ⅰ)连接 OG ,∵ EF 为 e O 的切线, ∴ OG ? EF , ∴ ?OGA ? ?KGE ? 90 ,
0

A H O K D

∵ CD ? AB ,∴ ?OAG ? ?HKA ? 90 ,
0

B

F G E ∵ OA ? OG ,∴ ?OGA ? ?OAG , ∴ ?KGE ? ?HKA ? ?GKE ,∴ KE ? GE . ······································································· 5 分

KG GE KE , ? ? KD KG KG ∵ ?DKG ? ?GKE ,∴△ KDG ∽△ KGE ∴ ?AGD ? ?E ,又∵ ?AGD ? ?ACD ,∴ ?ACD ? ?E .
(Ⅱ)连接 DG , ,∵ KG ? KD gGE ,∴
2

∴ AC P EF .

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

23.解: (I)圆 C1 的普通方程为: ( x ? 4) ? y ? 16 ,则 C1 的极坐标方程为: ? ? 8cos ?
2 2

圆 C2 的普通方程为: x ? ( y ? 2) ? 4 ,则 C2 的极坐标方程为: ? ? 4sin ?
2 2

··········· 5 分

(II) 设 P (?, ?) , 则 有 8cos ? ? 4sin ? , 解 得 tan ? ? 2 , sin ? ?

2 5 ,所以 P 点的极坐标为 5

(

8 5 2 5 , arcsin ) 5 5

························································································································································ 10 分 24.解: ( I)

5 1 ? 3 ?1 ?? x ? ? 2 ? x ? ? 2 原不等式等价于 ? 2 或 ?2 或 2 2 ? ? x ?1 ? ? 1? x ? 3

5 ?3 ? x? ?2 2 ?2 ? ? x?3

解得原不等式解集为 ( ??, ) U (3, ??) ····················································································· 5 分

1 3

5 ? 3 ? x ? , x ?1 ? 2 2 ? 1 1 ?1 (II) f ( x) ? x ? 1 ? | x ? 3 |? ? x ? ,1 ? x ? 3 2 2 ?2 5 ? 3 ? 2 x? 2,x ?3 ? f ( x) 图象如图所示,其中 A(1,1), B (3, 2) ,
直线 y ? a ( x ? ) 绕点 ( ?

············································· 7 分

y

1 , 0) 旋转, 2 x O 1 3 1 由图可得不等式 f ( x) ? a ( x ? ) 的解集非空时, a 的范围为 2 3 4 ············································································································ 10 分 ( -?, - ) U[ , +?) 2 7

1 2

A

B


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以上会员名单排名不分前后