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定义数列?an?： a1?1,a2?2，且对任意正整数n，有
nn?1?an?2??2?(?1)a?(?1)?1. n??（1）求数列?an?的通项公式与前n项和Sn；
（2）问是否存在正整数m,n，使得S2n?mS2n?1？若存在，则求出所有的正整数对
(m,n)；若不存在，则加以证明.
2012年深圳市高三年级第二次调研考试
数学（文科）参考答案及评分标准2012-4-23
说明：
1. 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．
2. 对计算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． 4. 只给整数分数，选择题和填空题不给中间分数．
一、选择题：本大题考查基本知识和基本运算。共10小题，每小题5分，满分50分． 题号 答案 1 C 2 C 3 A 4 D 5 B 6 A 7 B 8 C 9 C 10 D
二、填空题：本大题考查基本知识和基本运算，体现选择性．共5小题，每小题5分，满
分20分．其中第14、15两小题是选作题，考生只能选做一题，如果两题都做，以第14题的得分为最后得分． 11．4, 3（第一空3分，第二空2分） 12．2 13.
1722 14．1 15． 9903三、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． 16．（本小题满分12分）
在?ABC中，角A为锐角,记角A,B,C所对的边分别为a,b,c.设向量
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πm?(cosA,sinA),n?(cosA,?sinA),且m与n的夹角为.
3（1）求m?n的值及角A的大小；
（2）若a?7,c?3，求?ABC的面积S．
【说明】 本小题主要考查向量的数量积和夹角的概念，以及用正弦或余弦定理解三角形，
三角形的面积公式，考查了简单的数学运算能力．
解：（1）?m?cos2A?sin2A?1,n?cos2A?(?sinA)2?1,
?cos? ?m?n=m?nπ31.· ··········································································· 3分 2?m?n=cos2A?sin2A?cos2A，
1?cos2A?. ································································································· 5分
2π?0?A?,0?2A?π,
2ππ?2A?,A?. ························································································· 7分
36π222（2）(法一) ?a?7,c?3,A?,及a?b?c?2bccosA,
6?7?b2?3?3b, 即b??1(舍去)或b?4. ···································· 10分 1bcsinA?3. ······································································· 12分 2πac?(法二) ?a?7,c?3,A?,及,
6sinAsinC故S??sinC?csinA3. ····································································· 7分 ?a27?a?c, π5?0?C?,cosC?1?sin2A? 227π132 ?sinB?sin(π?A?C)?sin(?C)?cosC?sinC?6227asinB?b??4. ············································································· 10分
sinA1 故S?bcsinA?3.······································································ 12分
2
17．（本小题满分12分）
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设函数f(x)?x2?bx?c，其中b,c是某范围内的随机数，分别在下列条件下，求事件A “f(1)?5且f(0)?3”发生的概率. (1) 若随机数b,c?{1,2,3,4}；
(2) 已知随机函数Rand()产生的随机数的范围为x0?x?1, b,c是算法语句
??b?4?Rand()和c?4?Rand()的执行结果．(注: 符号“?”表示“乘号”)
【说明】本题主要考查随机数、随机函数的定义，古典概型，几何概型，线性规划等基
础知识，考查学生转换问题的能力，数据处理能力． 解：由f(x)?x2?bx?c知，事件A “f(1)?5且f(0)?3”，即??b?c?4···· 1分 . ·
?c?3 (1) 因为随机数b,c?{1,2,3,4}，所以共等可能地产生16个数对(b,c)，
列举如下：
(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(3,4)， (4,1),(4,2),(4,3),(4,4). ············································································· 4分
?b?c?4事件A ：?包含了其中6个数对(b,c)，即：
c?3?(1,1),(1,2),(1,3),(2,1),(2,2),(3,1). ······························································ 6分
所以P(A)?633?，即事件A发生的概率为. ········································· 7分 1688(2) 由题意，b,c均是区间[0,4]中的随机数，产生的点(b,c)均匀地分布在边长为4的正方形区域?中（如图），其面积S(?)?16. ····························· 8分
事件A ：??b?c?4所对应的区域为如图所示的梯形（阴影部分），
?c?3115?(1?4)?3?. ······················································· 10分 22c43(1,3)其面积为：S(A)?
15S(A)215??所以P(A)?， S(?)1632
即事件A的发生概率为
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O4b15. ······································································ 12分 32通达教学资源网 http://www.nyq.cn/
18．（本小题满分14分）
ABCD是平行四边形，E,F分别在棱BB1, 如图，四棱柱ABCD?A1BC11D1的底面
DD1上，且AF?EC1．
（1）求证：AE?FC1；
（2）若AA1?平面ABCD，四边形AEC1F是边长为6的正方形，且BE?1，DF?2，
求线段CC1的长, 并证明：AC?EC1.
DAD1FB1A1CEC1C1D1FA1DBB1O1COAEB第18题图
【说明】本题主要考察空间点、线、面位置关系，考查线线、线面平行的性质和判定，线线
垂直的性质和判定，考查空间想象能力、运算能力、把空间问题转化为平面问题的意识以及推理论证能力． 证明：（1）?四棱柱ABCD?A1BC11D1的底面ABCD是平行四边形，
·········································································· 1分 ?AA1?DD1,AB?CD. ·
?DD1,CD?平面CDD1C1,AA1,AB?平面CDD1C1,
········································· 3分 ?AA1?平面CDD1C1, AB?平面CDD1C1, ·
?AA1,AB?平面ABB1A1,AA1?AB?A，
································································ 4分 ?平面ABB1A1?平面CDD1C1. ·
?AF?EC1，
············································································ 5分 ?A,E,C1,F四点共面. ·
?平面AEC1F?平面ABB1A1?AE,平面AEC1F?平面CDD1C1?FC1,
······························································································ 7分 ?AE?FC1. ·
（2） 设AC?BD?O,AC1?EF?O1,
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? 四边形ABCD,四边形AEC1F都是平行四边形，
?O为AC，BD的中点，O1为AC1，EF的中点. ································ 8分
连结OO1,由(1)知BE?DF,从而OO1?11CC1?(BE?DF). 22?BE?1，DF?2，
································································································ 10分 ?CC1?3. ·
?AA1?平面ABCD,四边形AEC1F是正方形, ??ACC1，?ABE，?ADF均为直角三角形,得 AC2?AC12?CC12?2AE2?CC12?12?9?3, AB2?AE2?BE2?6?1?5,
BC2?AD2?AF2?DF2?6?4?2.
?AC2?BC2?AB2?5，即AC?BC. ·············································· 12分
?BB1?平面ABCD,AC?平面ABCD, ?AC?BB1.
?BC,BB1?平面BB1C1C,
?AC?平面BB1C1C. ·············································································· 13分
?EC1?平面BB1C1C,
··························································································· 14分 ?AC?EC1. ·
19．（本小题满分14分）
已知二次函数f?x?的最小值为?4,且关于x的不等式f?x??0的解集为
?x?1?x?3,x?R?,
（1）求函数f?x?的解析式；
f?x??4lnx的零点个数. （2）求函数g(x)?x【说明】本题主要考查二次函数与一元二次不等式的关系，函数零点的概念，导数运算
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