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广东省海珠区2012-2013学年高二下学期学期期末数学(理)试题(word版,含答案)

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海珠区 2012 学年第二学期期末考试试题 高二数学(理科)
本试卷共 4 页,20 小题,满分 150 分.考试用时 120 分钟. 注意事项: 1.答卷前,考生务必用黑色字迹钢笔或签字笔将自己的姓名和考生号填写在答题卡指定的 位置上. 2.选择题每小题选出答案后,用 2B 铅笔把答题卡上对应题目的答案标号涂黑;如需改动, 用橡皮擦干净后,再选涂其它答案,答案不能答在试卷上. 3.非选择题必须用黑色字迹的钢笔或签字笔作答,答案必须写在答题卡各题目指定区域内 的相应位置上;如需改动,先划掉原来的答案,然后再写上新的答案;不准使用铅笔和 涂改液.不按以上要求作答的答案无效. 4.本次考试不允许使用计算器. 5.考生必须保持答题卡的整洁,考试结束后,将试卷和答题卡一并交回. 公式与数据: 1.在 n 次独立重复试验中,用 X 表示事件 A 发生的次数,若每次试验中事件 A 发生的概率 为 p,则

一、选择题:本大题共 8 小题,每小题 5 分,满分 40 分.在每小题给出的四个选项中,只 有 一项是符合题目要求的.

1.若 i 为虚数单位,则复数

在复平面内对应的点所在象限为

A.第一象限 B.第二象限 C.第三象限 D.第四象限 2.命题“所有能被 2 整除的数都是偶数”的否定是 A.存在一个能被 2 整除的数不是偶数 B.存在一个不能被 2 整除的数是偶数 C.所有不能被 2 整除的数都是偶数 D.所有能被 2 整除的数都不是偶数 3.已知随机变量 x 服从正态分布 A.0.6 4.由曲线 B.0.4 C.0.3 ,且 D.0.2

所围成图形的面积为

A.

B.

C.

D.

5.双曲线

的实轴长是

A. 6.若 A.

B.4

C. ,则

D.2 的解集为( ) C. D.

B.

7. 某外商计划在四个候选城市投资 3 个不同的项目,且在同一个城市投资的项目不超过 2 个, 则该外商不同的投资方案有( ) A.60 种 B.42 种 C.36 种 D. 16 种 8.设 f ( x) 是定义在正整数集上的函数,且 f ( x) 满足: “当 成立” .那么,下列命题总成立的是( ) A.若 f(1) < 1 成立,则 f(10) <100 成立 B.若 f (2) < 4 成立,则 1 (1) f ? 成立 C.若 f (3)>=9 成立,则当 k>=1 时,均有 D.若 f (4)>=25 成立,则当 4 k? 时,均有 成立 成立 成立时,总可推出

二、填空题:本大题共 6 小题,每小题 5 分,满分 30 分. 9.已知向量 ,则 .

10.在

的二项展开式中,第 4 项的系数为 .

11.一位国王的铸币大臣在每箱 100 枚的硬币中各掺入了 10 枚劣币,国王怀疑大臣作弊, 他用在 10 箱子中各任意检查一枚的方法来检测,国王能发现至少一枚劣币的概率为 . 12.已知某生产厂家的年利润 y(单位:万元)与年产量 x(单位:万件)的函数关系式为

,则使该生产厂家获得最大年利润的年产量为 万件.

13.设函数

,已知

,

,

,

, 根据以上事实,由归纳推理可得: 当 ,且 时, = . 的焦点为 F,点 A(0, 2) ,若线段 FA 的中点 B 在抛物线上,则

14. 抛物线

B 到该抛物线焦点的距离为



三、解答题:本大题共 6 小题,满分 80 分.解答须写出文字说明、证明过程和演算步骤. 15. (本小题满分 12 分)

求函数

的极值.

16.(本小题满分 12 分) 已知动点 M 在直线 l : y= 2 的下方,点 M 到直线 l 的距离与到定点 N(0, -1)的距离之和为 4,求动点 M 的轨迹方程.

17. (本小题满分 14 分) 设 (1)求 的值; (2)证明:对任意实数 c,直线 与函数 的图象不相切. , 图像的一条对称轴是 .

18.(本小题满分 14 分) 如图,在梯形 ABCD 中,AB/ / CD,AD=DC=CB=1, ,四边形 ACFE 为

矩形,平面 平面 ABCD,CF= 1. (1)求证: 平面 ACFE; (2)若点 M 在线段 EF 上移动,试问是否存在点 M,使得平面 MAB 与平面 FCB 所成的二 面角为 ,若存在,求出点 M 的坐标;若不存在,说明理由.

19. (本小题满分 14 分) PM2.5 是指大气中直径小于或等于 2.5 微米的颗粒物,也称为可入肺颗粒物.我国 PM2.5 标 准采用世卫组织设定的最宽限值, PM2.5 日均值在 35 微克/立方米以 即 下空气质量为一级;在 35 微克/立方米~75 微克/立方米之间空气质量 为二级;在 75 微克/立方米以上空气质量为超标. 某城市环保局从该市市区 2012 年全年每天的 PM2.5 监测 数据中随机的抽取 15 天的数据作为样本,监测值如茎叶图所示 (十位为茎,个位为叶) . (1)从这 15 天的 PM2.5 日均监测数据中,随机抽出三天数 据,求恰有一天空气质量达到一级的概率; (2)从这 15 天的数据中任取三天数据,记 表示抽到 PM2.5 监测数据超标的天数,求 的分布列和数学期望; (3)根据这 15 天的 PM2.5 日均值来估计一年的空气质量情 况,则一年(按 365 天计算)中平均有多少天的空气质量达到 一级或二级?

20.(本小题满分 14 分)

如图,已知椭圆 的方程为

,圆 的方程为

,斜率为

的直线 过椭圆 的左顶点 A,且与椭圆 、圆 分别相交于 B,C. (1)若 = 1 时,B 为线段 AC 的中点,求椭圆 的离心率 e; 为椭圆的右焦点,当 时,求 的值;

(2)若椭圆 的离心率 e= ,

(3)设 D 为圆 上不同于 A 的一点,直线 AD 的斜率为 ,当 否过定点?若过定点,求出定点坐标;若不过定点,请说明理由.

时,直线 BD 是

海珠区 2012 学年第二学期期末考试

高二数学(理科) 参考答案及评分标准
说明: 1.本解答给出了一种解法供参考, 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.
[来源:Zxxk.Com]

2.对于解答题, 当考生的解答在某一步出现错误时, 如果后续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给分,但不得超过该部分正确解答应得分数的一 半;如果后续部分的解答有较严重的错误,就不再给分. 3.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. 一、选择题:本大题共 8 小题,每小题 5 分,满分 40 分. 题号 答案 1 D 2 A 3 D 4 C 5 B 6 B 7 A 8 D

二、填空题:本大题共 6 小题,每小题 5 分,满分 30 分. 9. 13; 10. ?40 ; 11. 0.651 ; 12.9; 13.

x ; (2 ? 1) x ? 2n
n

14.

3 2. 4
三、解答题:本大题 共 6 小题,满分 80 分.解答须写出文字说明、证明过程和演算步骤. 15. (本小题满分 12 分)

1 3 x ? 9 x ? 1( x ? R ) 的极值. 3 1 3 2 【解】因为 f ( x ) ? x ? 9 x ? 1( x ? R ) ,所以 f '( x) ? x ? 9 ? ( x ? 3)( x ? 3) ......................... 2 分 3
求函数 f ( x ) ? 令 f '( x) ? 0 ,解得 x ? ?3 ,或 x ? 3 . ........................................................................................ 3 分 由 f '( x) ? 0 ,得 x ? ?3 ,或 x ? 3 ;由 f '( x) ? 0 ,得 ?3 ? x ? 3 . ································ 4 分 当 x 变化时, f '( x) , f ( x ) 的变化情况如下表:zxxk

x
f '( x)

(??, ?3)

?3

(?3,3)

3

(3, ??)

?

0

?

0

?

f ( x)

单调递增 ?

19

?17

单调递减 ?

单调递增 ?

(单调区间列,每列对一列记 1 分;两个极值点值,全列对记1分) ································ 8 分 因此当 x ? ?3 时, ··························································································································· 9分

f ( x) 有极大值,极大值为 f (?3) ? 19 ; ··················································································· 10 分
当 x ? 3 时, ······································································································································ 11 分

f ( x) 有极小值,极小值为 f (3) ? ?17 . ···················································································· 12 分
16.(本小题满分 12 分) 已知动点 M 在直线 l : y ? 2 的下方, M 到直线 l 的距离与定点 N (0, ?1) 的 距离之和 点 为 4,求动点 M 的轨迹方程. 【解】设动点 M 的坐标为 M ( x, y ) .························································································· 1 分 因为点 M 在直线 l : y ? 2 的下方,所以 y ? 2 ,依题意有

x 2 ? ( y ? 1) 2 ? | y ? 2 |? 4 ........................................................................................................... 4 分
(评 分说明:两个距离写对一个即计 1 分,写对两个记2分.)
2 2 因为 y ? 2 ,所以 x ? ( y ? 1) ? y ? 2 .................................................................................... 6 分

1 2 ( x ? 3) , .......................................................................................................... 8 分 2 1 2 因为 y ? 2 ,所以 ( x ? 3) ? 2 ,............................................................................................... 9 分 2
平方化简得 y ? 解得 ? 7 ? x ? 7 , ..................................................................................................................... 10 分 所以所求的轨迹方程为 y ?

1 2 ( x ? 3)(? 7 ? x ? 7) .................................................................. 12 分 2

(评分说明:未注明 x 的取值范围或取值范围错误均要扣 1 分.) 17. (本小题满分14分) 设 f ( x) ? sin(2 x ? ? )(?? ? ? ? 0), f ( x) 图像的一条对称轴是 x ? (1)求 ? 的值;

?
8

.

(2)证明:对任意实数 c ,直线 5x ? 2 y ? c ? 0 与函数 y ? f (x) 的图象不相切.

【解析】 (1)由对称轴是 x ?

?
8

,得 sin(

?
4

? ? ) ? ?1 , ························································ 2 分

(评分说明: ?1 处遗漏 一个记 1 分.) 即

?
4

? ? ? k? ?

?
2

. ························································································································ 3 分

所以 ? ? k? ?

?
4

? k ? Z ? . ··············································································································· 4分

3 4 3 (2)因为 f ( x) ? sin(2 x ? ? ) . 4 3 所以 f ?( x) ? 2 cos(2 x ? ? ) , ···································································································· 8 分 4
而 ?? ? ? ? 0 ,所以 ? ? ? ? . ································································································ 6 分

? 2 ································································································································· 10 分
而直线 5x ? 2 y ? c ? 0 的斜率 k ?

5 ? 2 , ··············································································· 12 分 2

所以直线 5x ? 2 y ? c ? 0 不是 函数 y ? f (x) 的切线. ··························································· 14 分 18.(本小题满分 14 分) 如图, 在梯形 ABCD 中, AB / / CD ,AD ? DC ? CB ? 1 ,?ABC ? 60o , 边形 ACFE 四 为矩形,平面 ACFE ? 平面 ABCD , CF ? 1 . (1)求证: BC ? 平面 ACFE ; (2)若点 M 在线段 EF 上移动,试问是否存在点 M ,使得平面 MAB 与平面 FCB 所 成的二面角为 45o ,若存在,求出点 M 的坐标;若不存在,说明理由.

F
M

E

D

C

B A
【解】 (1)证明:在梯形 ABCD 中, AB / / CD , AD ? DC ? CB ? 1 , ?ABC ? 60o , 所以 AB ? 2 , ·································································································································· 1分

AC 2 ? AB 2 ? BC 2 ? 2 AC ? BC cos 60? ? 3 , ·········································································· 2 分
所以 AB 2 ? AC 2 ? BC 2 ,

所以 AC ? BC , ····························································································································· 3 分 又平面 ACFE ? 平面 ABCD , AC 是交线, BC ? 平面 ABCD ,········································ 4 分 所以 BC ? 平面 ACFE . ··················································································································· 5 分

AC (2) (1) AC , BC , CF 两两互相垂直, C 为坐标原点, , BC , CF 分别为 x, y , z 由 知 以
轴建立空间直角坐标系,如图, ··········································································································· 6 分 则 A, B 的坐标分别为 A( 3,0,0), B(0,1,0) ,设 M 的坐标为 M (a,0,1) ,则

??? ? ???? ? AB ? (? 3,1,0), BM ? (a, ?1,1) , ···························································································· 7 分
设 m ? ( x, y, z) 是平面 AMB 的法向量,则

??

?? ??? ? ?m ? AB ? ? 3x ? y ? 0 ? (评分说明:列对一个各记 1 分) ··············································· 9 分 ? ? ?? ???? ?m ? BM ? ax ? y ? z ? 0 ?

F
M

E

D

C

B A



x ?1





?? m?(

?a 1

, ,

3

,

···································································································································································· 10 分 显 然

? n?(

1

, 是 0 平,

面 0

)F

C 的 B









···································································································································································· 11 分





?? ? cos ? m, n ??

1 4 ? ( 3 ? a) 2

?

2 2

.

···································································································································································· 12 分 化 简 得

2 ? ( 3 ? a)2 ? 0



















···································································································································································· 13 分 所 以 线 段 EF 上 不 存 在 点 M 使 得 平 面 MAB 与 平 面 FCB 所 成 二 面 角 为 45? . ···································································································································································· 14 分 19. (本小题满分14分)

PM2.5 是指大气中直径小于或等于 2.5 微米的颗粒物,也称为可入肺颗粒物.我国 PM2.5 标准采用世卫组织设定的最宽限值,即 PM2.5 日均值在 35 微克/立方米以下空 气质量为一级;在 35 微克/立方米 ? 75 微克/立方米之间空气质量为二级;在 75 微克/
立方米以上空气质量为超标. 某城市环保局从该市市区 2012 年全年每天的 PM2.5 监测数 据中随机的抽取 15 天的数据作为样本, 监测值如茎叶图所 示 (十位为茎,个位为叶) . (1)从这 15 天的 PM2.5 日均监测数据中,随机抽出三 天数据, 求恰有一天空气质量达到一级的概率;
] [来源:学科网]

PM2.5 日均值
(微克/立方米)

2 3 4 6

8 7 4 3

[

1 5 8

4 5

3
[ 来

源:Z.xx.k.

Com]

(2)从这 15 天的数据中任取三天数据,记 ? 表示抽到

PM2.5
监测数据超标的天数,求 ? 的分布列和数学期望;

7 8 9

9 6 2

3 5

(3)根据这 15 天的 PM2.5 日均值来估计一年的空气质量情况,则一年(按 365 天计 算)中平均有多少天的空气质量达到一级或二级. 【解】 (1)从茎叶图可知,空气质量为一级的有 4 天,为二级 的有 6 天,超标的有 5 天, ···························································································································································

1分 记 15 天的 PM2.5 日均监测数据中, “ 随机抽出三天数据, 恰有一天空气质量达到一级” 为事件 A , ··························································································································································· 2分 则

P( A) ?

2 C1 ? C11 44 4 ? 3 C15 91



···································································································································································· 4分 (2) ? 的可能值为 0,1,2,3.

P (? ? 0) ?

0 3 C1 C2 45 C5 C10 24 , P(? ? 1) ? 5 3 10 ? , ? 3 C15 91 C15 91

P (? ? 2) ?

1 C52C10 20 C 3C 0 2 , P (? ? 3) ? 5 3 10 ? .(评分说明:对一个各记 1 分) ? 3 C15 91 C15 91

···································································································································································· 8分 所以 ? 的分布列为

?

0

1

2

3

P
( 评 分

24 91
说 明 : 列

45 91
对 两 个

20 91
各 记

2 91
1 分 )

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E? ?

24 45 20 2 ? 0 ? ?1 ? ? 2 ? ? 3 91 91 91 91



····································································································································································

11 分

?1
···································································································································································· 12 分 ( 3 ) 15 天 的 空 气 质 量 达 到 一 级 或 二 级 的 频 率 为

10 2 2 1 ? , 365? ? 243 , 15 3 3 3

···································································································································································· 13 分 所 以 估 计 一 年 中 有 243 天 空 气 质 量 达 到 一 级 或 二 级 .

···································································································································································· 14 分 (说明:答 243 天,244 天不扣分) 20.(本题满分 14 分) 如图,已知椭圆 E1 的方程为

x2 y 2 ? 2 ? 1(a ? b ? 0) ,圆 E2 的方程为 x 2 ? y 2 ? a 2 , 2 a b

斜率为 k1 的直线 l1 过椭圆 E1 的左顶点 A ,且与椭圆 E1 、圆 E2 分别相交于 B , C .
y

(1)若 k1 ? 1 时, B 为线段 AC 的中点,求椭圆 E1 的 离心率 e ;
D A B

C

1 (2)若椭圆 E1 的离心率 e = , F2 为椭圆的右焦点, 2
当 值;

O

l1

x

l2

| BA | ? | BF2 |? 2a 时,求 k1 的

E1

E2

k1 b 2 ? (3) D 为圆 E2 上不同于 A 的一点, 设 直线 AD 的斜率为 k2 , 当 时, 直线 BD k2 a 2
是否过定点?若过定点,求出定点坐标;若不过定点,请说明理由. 【解】 (1)当 k1 ? 1 时,点 C 在 y 轴上,且点 C 的坐标为 C (0, a ) ,则点 B 的坐标为

? a a? B ? ? , ?, ? 2 2?

··························································································································································· 1分





B













a a (? ) 2 ( ) 2 2 ? 2 ?1 a2 b2



···································································································································································· 2分 所 以

b2 1 ? a2 3



···································································································································································· 3分

e2 ?

c2 b2 2 ? 1? 2 ? a2 a 3







e?

6 3

.

···································································································································································· 4分 (2)设椭圆的左焦点为 F ,由椭圆的定义知 | BF | ? | BF2 |? 2a , 1 1 所 以

|B

1

?

F|



|B

···································································································································································· 5分 即

B



线



AF1







线









xB ? ?

a?c 2



···································································································································································· 6分 又因为 e ?

c 1 1 3 ? ,所以 c ? a, b ? a, a 2 2 2

[来源:学+科+网 Z+X+X+K]

所 以

3 xB ? ? a , 代 入 4

椭 圆 方 程 得

yB ? ?

7 21 b?? a , 4 8

···································································································································································· 7分
[来源:Zxxk.Com]





k1 ?

yB 21 ?? xB ? a 2

.

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



3











? y ? k1 ( x ? a ), ? 2 ?x y2 ? 2 ? 2 ? 1, b ?a

···································································································································································· 9分

x 2 ? a 2 k12 ( x ? a ) 2 得 ? ? 0, a2 b2
所 以

x ? ?a





x?

a (b 2 ? k12a 2) b 2 ? a 2k12



···································································································································································· 10 分 所 以 xB ? ? a , 所 以 xB ?

a (b 2 ? k12 a 2 ) 2ab 2 k1 , 则 yB ? k1 ( xB ? a ) ? 2 . b 2 ? a 2 k12 b ? a 2 k12

···································································································································································· 11 分 由?

? y ? k2 ( x ? a) ?x ? y ? a
2 2 2

2 ,得 x2 ? a2 ? k2 ( x ? a)2 ? 0

得 x ? ?a ,或 x ?

2 a(1 ? k2 ) , 2 1 ? k2 2 a(1 ? k2 ) 2ak2 , yD ? 2 2 1 ? k2 1 ? k2





xD ?



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

b4 2 k2 ) 2 a ( a 2 ? b 2 k2 ) 2ab 2 k2 k1 b a2 当 ? 2 2 2 , yB ? 2 2 2 , ? 时, xB ? b4 2 a ? b k2 a ? b k2 k2 a 2 2 b ? 2 k2 a
2

a(b2 ?

k BD

2ab 2 k2 2ak2 ? 2 2 2 2 a ? b k2 1 ? k2 1 ? ?? 2 2 2 2 a(a ? b k2 ) a(1 ? k2 ) k2 ? 2 2 2 2 a ? b k2 1 ? k2



···································································································································································· 13 分 所以 BD ? AD . 因 为 E2 为 圆 , 所 以 ? ADB 所 对 圆 E2 的 弦 为 直 径 , 从 而 BD 经 过 定 点 ( a, 0) . ···································································································································································· 14 分 法 二 : 直 线

BD







(a, 0)



···································································································································································· 6分 证明如下: 设

P(a, 0)



B ( xB , y B )





xB 2 y B 2 ? 2 ? 1(a ? b ? 0) a2 b



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

k AD k PB ?

y y y2 a2 a2 a2 a 2 b2 k1k PB ? 2 ? B ? B ? 2 ? 2 B 2 ? 2 (? 2 ) ? ?1 b2 b xB ? a xB ? a b x B ? a b a



····································································································································································

10 分 所 以

P B? A D



···································································································································································· 12 分 又 PD ? AD , 所 以 三 点 P, B, D 共 线 , 即 直 线 BD 过 定 点 P (a, 0) . ···································································································································································· 14 分

zxxk


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