2012大连-沈阳高三联合第一次模拟考试（大连一模数学文科答案）

2012年大连-沈阳联合模拟考试
文科数学试题参考答案
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．D ；3．C ；4．A ；5．C ；6．B ；7．C ；8．B ；9．C ；10．A ；11．B ； 12．D ． 二、填空题
13．3-；14．1
3,(1)
23.(2)
n n n -=ìí·³î；15． 29p ；16．(,1)-¥． 三、解答题
17．解：(Ⅰ)由频率分布表得a ＋0.2＋0.45＋b ＋c ＝1，即0.35a b c ++=. ·················· 2分
因为抽取的20件轴承中，等级编号为4的恰有3件，所以3
0.1520b ==. 等级编号为5的恰有2件，所以2
0.120
c =
=． ·
························································· 4分 从而0.350.1a b c =--=．
所以0.1a =，0.15b =，0.1c =． ············································································· 6分
(Ⅱ)从轴承123,,x x x ，12,y y 中任取两件，所有可能的结果为：中任取两件，所有可能的结果为： {}{}{}{}{}{}121311122321,,,,,,,,,,,,x x x x x y x y x x x y
{}{}{}{}22313212,,,,,,,x y x y x y y y ． ·
·········································································· 8分 设事件A 表示“从轴承123,,x x x ，12,y y 中任取两件，其等级编号相等”，则A 包含的基本事件为：的基本事件为：
{}{}{}{}12132312,,,,,,,x x x x x x y y 共4个．个． ······························································· 10分 又基本事件的总数为10， 故所求的概率4
()0.410
P A ==．
················································································ 12分
1313 )
p
p
)
7]
p p p p
p
)
]
11
2012年大连市高三一模文科数学试题参考答案与评分标准年大连市高三一模文科数学试题参考答案与评分标准
∴b x
x x
bx x f ³-
+
Û-³ln 112)(，················································································
8分 令x x x x g ln 11)(-+=，可得)(x g 在(]2,0e 上递减，在[)
+¥,2e 上递增，上递增， ····················· 10分 ∴22min 11)()(e
e g x g -
==，即211b e £-． ································································· 12分 21．解：（Ⅰ）∵点M 到抛物线准线的距离为=+2
4p 4
17，
∴21=p ，即抛物线C 的方程为x y =2．·································································· 2分 （Ⅱ）法一：∵当AHB Ð的角平分线垂直x 轴时，点)2,4(H ，∴HE HF k k =-， 设11(,)E x y ，22(,)F x y ，
∴1212H H H H y y y y x x x x --=---，∴，∴ 122222
12
H H H H y y y y y y y y --=---， ∴1224H y y y +=-=-. ·························································································· 5分
2121
2
2
21
2121
11
4
EF y y y y k x x y y y y --=
=
=
=---+．
································································ 7分 法二：∵当AHB Ð的角平分线垂直x 轴时，点)2,4(H ，∴
60=ÐAHB ，
可得3=H A
k ，
3-=H B k ，∴直线HA 的方程为2343+-=x y ，
联立方程组î
íì=+-=x y x y 22343，得023432=+--y y ， ∵323
E y +=
∴36
3-=E
y ，3
3413-=E x ． ··········································································· 5分 同理可得363--=
F y ，3
3413+=F x
，∴41-=EF k ．···································
7分
11
(15-
15
m -(
所以DE CB
CE AB =
，所以2BC =． ············································································· 10分 23．解：．解：((Ⅰ)2cos ,2sin 2.
x y
a a =
ìí=+î 且参数[]0,2a p Î， 所以点P 的轨迹方程为22(2)4x y +-=． ···························································· 3分
(Ⅱ)因为)
4
sin(210p q r -=，所以2sin()104
p
r q -=，
所以sin cos 10r q r q -=，所以直线l 的直角坐标方程为100x y -+=． ········· 6分 法一：由法一：由((Ⅰ) ) 点点P 的轨迹方程为22
(2)4x y +-=，圆心为(0,2)，半径为2.
22101210
4211
d ´-´+==+，所以点P 到直线l 距离的最大值422+. ············ 10分
法二：222cos 2sin 21022cos()44
11d a a p a --+==+++，当74p a =，max 422d =+，即点P 到直线l 距离的最大值422+. ·
·································· 10分 24．解：（Ⅰ）由26x a a -+£得26x a a -£-，∴626a x a a -£-£-，
即33a x -££，∴32a -=-，∴1a =． ································································· 5分
（Ⅱ）由（Ⅰ）知()211f x x =-+,令()()()n f n f n j =+-，
则()124, 211
212124, 221
24, n 2n n n n n n n j ì-£-ïï
ï
=-+++=-<£íï
ï+>
ïî
∴()n j 的最小值为4，故实数m 的取值范围是[
)4,+¥． ········································· 10分。