福州市第二次质检数学文试题及答案(3月)

2015年福州市高中毕业班质量检测
文科数学能力测试
（完卷时间：120分钟；满分：150分）
注意事项：
1．本科考试分试题卷和答题卷，考生须在答题卷上作答，答题前，请在答题卷的密封线内填写学校、班级、准考证号、姓名；
2．本试卷分为第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分，全卷满分150分，考试时间120分钟．
参考公式：
1．样本数据12,,
,n x x x 的标准差
()()()22
2
121n s x x x x x x n ⎡⎤=
-+-++-⎣
⎦
，
其中x 为样本平均数；
2．柱体体积公式：V Sh =， 其中S 为底面面积，h 为高； 3．锥体体积公式：1
3
V Sh =
， 其中S 为底面面积，h 为高.
第Ⅰ卷 （选择题 共60分）
一、选择题（本大题共12小题，每小题5分，共60分．在每小题所给的四个选项中有且
只有一个选项是正确的．把正确选项涂在答题卡的相应位置上．） 1． 函数1y x =-的定义域为
A ．{}1x x
B ．{}1x x <
C ．{}1x x
D ．{}1x x >
2． 若复数z 满足()2i i z -⋅=（i 为虚数单位），则z 的虚部为
A ．
25 B ．25- C ．15 D ．15
- 3． 如图是某篮球联赛中，甲、乙两名运动员12个场次得分的茎叶图．设甲、乙两人得
分的平均数分别为x 甲,x 乙，中位数分别为m 甲,m 乙，则 A ．,x x m m <<甲乙甲乙 B ．,x x m m <>甲乙甲乙 C ．,x x m m >>甲乙甲乙 D ．,x x m m ><甲乙甲乙
4． 已知直线2y x =为双曲线2
2
22:
1x y a b Γ-=（0,0a b >>）的一条渐近线，则双曲线Γ的离心率为 A ．
3 B ．5 C ．2 D ．5
第3题图
5． 执行如图所示的程序框图，输出的有序实数对为
A ．()8,2
B ．()8,3
C ．()16,3
D ．()16,4
6． 已知直线l 与平面α平行，则下列结论错误．．
的是 A ．直线l 与平面α没有公共点 B ．存在经过直线l 的平面与平面α平行 C ．直线l 与平面α内的任意一条直线平行 D ．直线l 上所有的点到平面α的距离都相等
7． 已知偶函数()f x 满足：当()12,0,x x ∈+∞时，()()()12120x x f x f x ⎡⎤-->⎣⎦恒成立．设
(4)a f =-,(1)b f =,(3)c f =，则,,a b c 的大小关系为 A ．a b c << B ． b a c << C ．b c a << D ． c b a <<
8． 设变量,x y 满足约束条件,2,26,y x x y x y ⎧⎪
+⎨⎪+⎩
则32z x y =+的取值范围为
A ．(],10-∞
B ．[)8,+∞
C ．[]5,10
D ．[]8,10
9． 某四棱锥的三视图如图所示，该四棱锥的表面积为
A ．17
B ．22 C
．14+ D
．22+
10．函数22sin 10,()240
x x x f x x x x ⎧-+>⎪=⎨--⎪⎩，，的零点个数为 A ．0 B ．1 C ．2 D ．3 11．在ABC ∆中，点G 为ABC ∆的重心．
已知AB =，且向量GA 与GB 的夹角为120︒，
则CA CB ⋅的最小值是 A ．3-
B ．6
C ．9
D ．24
12．已知函数()sin f x x x =⋅，有下列三个结论：
①存在常数0T >，对任意的实数x ，恒有()()f x T f x +=成立； ②对任意给定的正数M ，都存在实数0x ，使得()
0f x M ；
③直线y x =与函数()f x 的图象相切，且切点有无数多个． 则所有正确结论的序号是 A ．①
B ．②
C ．③
D ．②③
第5题图
第10题图
俯视图
3?
y 否
第Ⅱ卷 （非选择题 共90分）
二、填空题（本大题共4小题，每小题4分，共16分. 把答案填在答题卡的相应位置上．） 13．已知集合{}{}
21,0,1A B x x =-=∈=R ，则集合．．A
B 等于 ★★★ ．
14．已知函数()ln f x x =．若在区间()0,3e 上随机取一个数x ，则使得不等式()1f x 成立
的概率为 ★★★ ．
15．ABC ∆的内角A B C ,,所对的边分别是a b c ,,．若105,15B C =︒=︒，则
2cos15cos105a
b c ︒+︒
的值为 ★★★ ．
16．在各项均为正整数的单调递增数列{}n a 中，11a =，22a =，且
132112,k k k k a a a a +++⎛⎫⎛⎫
++= ⎪⎪⎝
⎭⎝⎭
*k ∈N ，则9a 的值为 ★★★ ．
三、解答题(本大题共6小题,共74分,解答应写出文字说明、证明过程或演算过程.) 17．（本小题满分12分）
已知函数()cos (0)f x x x ωωω=->的图象与直线2y =的相邻两个交点之间的 距离为π．
（Ⅰ）求函数()f x 的单调递增区间；
（Ⅱ）若223f α⎛⎫= ⎪⎝⎭，求πcos 23α⎛
⎫- ⎪⎝
⎭的值．
18．（本小题满分12分）
调查表明，中年人的成就感与收入、学历、职业的满意度的指标有极强的相关性．现 将这三项的满意度指标分别记为,,x y z ，并对它们进行量化：0表示不满意，1表示基本满意，2表示满意，再用综合指标w x y z =++的值评定中年人的成就感等级：若4w ，则
成就感为一级；若23w ，则成就感为二级；若01w ，则成就感为三级．为了了解
目前某群体中年人的成就感情况，研究人员随机采访了该群体的10名中年人，得到如下
结果：
（Ⅱ）从成就感等级为一级的被采访者中随机抽取两人，这两人的综合指标w 均为4的概率是多少？
19．（本小题满分12分）
如图，在长方体1111ABCD A B C D -中，12,4AB BC AA ===，P 为 线段11B D 上一点．
（Ⅰ）求证：AC BP ⊥；
（Ⅱ）当P 为线段11B D 的中点时，求三棱锥A PBC -的高．
20．（本小题满分12分）
小辉是一位收藏爱好者，在第1年初购买了价值为20万元的收藏品M ，由于受到收 藏品市场行情的影响，第2年、第3年的每年初M 的价值为上年初的1
2
；从第4年开始，每年初M 的价值比上年初增加4万元.
（Ⅰ）求第几年初开始M 的价值超过原购买的价值；
（Ⅱ）记n T （*n ∈N ）表示收藏品M 前n 年的价值的平均值，求n T 的最小值．
21．（本小题满分12分）
已知函数()e x
x
f x m
=
+,m ∈R ,e 2.71828=⋅⋅⋅为自然对数的底数． （Ⅰ）若1x =是()f x 的极值点，求m 的值； （Ⅱ）证明：当01a b <<<时，e e a b b a a b +<+．
22．（本小题满分14分）
如图，已知椭圆2222:1x y a b Γ+=（0a b >>）的离心率
e =左焦点和右顶点，且3AF =．
（Ⅰ）求椭圆Γ的方程；
（Ⅱ）过点F 作一条直线l 交椭圆Γ于,P Q 两点，点Q x 轴的对称点为Q '．若PF AQ '∥，求证：1
2
PF AQ '=
．
第19题图
1
A A
2015年福州市高中毕业班质量检测
文科数学能力测试参考答案及评分细则
一、选择题：本题共有12个小题，每小题5分，满分60分． 1．A 2．A 3．C 4．D 5．D 6．C 7．C 8．B 9．D 10．B 11．B 12．D 二、填空题：本大题共 4 小题，每小题 4 分，满分 16 分．
13．{}1- 14．1
3
15．2 16．55
三、解答题：本大题共6小题，共74分．
17． 本小题主要考查三角函数的图象与性质（对称性、周期性、单调性）、二倍角的余弦公式等基础知识，考查运算求解能力，考查数形结合思想、化归与转化思想、函数与方程思想．满分12分．
【解析】（Ⅰ）因为()cos (0,)f x x x x ωωω=->∈R ，
所以()2sin()6
f x x π
ω=-． ··········································································· 2分
所以max ()2f x =．
因为函数()f x 与直线2y =的相邻两个交点之间距离为π， 所以T π=， ····························································································· 3分
所以2ππω
=，解得2,ω=·
··········································································· 4分 所以()2sin(2)6f x x π
=-．
令222,2
6
2
k x k k π
π
π
ππ-
-+
∈Z ， ·
··························································· 5分 解得,6
3
k x
k k π
π
ππ-
+
∈Z ． ·
·································································· 6分 所以函数()f x 的单调递增区间是[,],63k k k ππ
ππ-
+∈Z ． ·································· 7分 （Ⅱ）由（Ⅰ）知，2sin()26f απα⎛⎫=- ⎪⎝⎭，因为2
23
f α⎛⎫= ⎪⎝⎭，
所以1
sin()63πα-=． ·················································································· 8分
所以cos 2cos236ππαα⎛⎫⎛
⎫-=- ⎪ ⎪⎝⎭⎝
⎭ ································································· 10分
212sin 6πα⎛
⎫=-- ⎪⎝⎭
····························································· 11分
7
9
=． ············································································· 12分 18．本小题主要考查概率、统计等基础知识，考查数据处理能力、运算求解能力、应用意识，考查必然与或然思想．满分12分． 【解析】（Ⅰ）计算10名被采访者的综合指标，可得下表：
由上表可知：成就感为三级（即01w ）的只有9A 一位，其频率为1
10
． ··········· 3分
用样本的频率估计总体的频率，可估计该群体中成就感等级为三级的人数有1
2002010
⨯=．
·············································································································· 5分 （Ⅱ）设事件A 为 “从成就感等级是一级的被采访者中随机抽取两人，他们的综合指标w 均为4”．由（Ⅰ）可知成就感是一级的（4w ）有：123568,,,,,A A A A A A ，共6位，从中随机抽取两人，所有可能的结果为：
{}{}{}{}{}{}{}{}{}{}{}1213151618232526283536,,,,,,,,,,,,,,,,,,,,,,
A A A A A A A A A A A A A A A A A A A A A A {}{}{}{}38565868,,,,,,,A A A A A A A A ，共15种． ·
·················································· 9分 其中综合指标4w =有：125,,A A A ，共3名，事件A 发生的所有可能结果为：
{}{}{}121525,,,,,A A A A A A ，共3种， ·
·························· 10分 所以3()15P A =
=1
5
．·
················································································ 12分 19．本小题主要考查直线与直线、直线与平面的位置关系，以及几何体的体积等基础知识，考查空间想象能力、推理论证能力、运算求解能力，考查数形结合思想、化归与转化思想．满分12分．
证明：（Ⅰ）连结BD ．
因为1111ABCD A B C D -是长方体，且2AB BC ==， 所以四边形ABCD 是正方形， ······································································· 1分 所以AC BD ⊥． ························································································ 2分 因为在长方体1111ABCD A B C D -中，
1BB ⊥平面ABCD ，AC ⊂平面ABCD ， 所以1AC BB ⊥． ··························································· 4分 因为BD ⊂平面11BB D D ，1BB ⊂平面11BB D D ， 且1BD BB B =，
所以AC ⊥平面11BB D D ． ················································ 5分 因为BP ⊂平面11BB D D ， 所以AC BP ⊥ ······························································· 6分 （Ⅱ）点P 到平面ABC 的距离14AA =，
ABC ∆的面积1
22
ABC S AB BC ∆=⋅⋅=， ·
······················
·················
····················· 7分 所以1118
24=333
P ABC ABC V S AA -∆=⋅=⨯⨯． ····
·························
·····
······················· 8分
在Rt △1BB P 中，114,BB B P ==所以BP =， ·····················
····················· 9分
同理CP =又=2BC ，所以PBC ∆的面积1
22
PBC S ∆=⨯． ·
·· 10分 设三棱锥A PBC -的高为h ，则
1
A A
因为A PBC P ABC V V --=，所以18
33
PBC S h ⋅=， ····················································· 11分
83
=
，解得h =
即三棱锥A PBC -
·································································· 12分
20．本小题主要考查数列、等差数列、等比数列、不等式等基础知识，考查运算求解能力、抽象概括能力、应用意识，考查函数与方程思想、化归与转化思想、分类与整合思想．满分12分．
解：（Ⅰ）设第n 年初M 的价值为n a ，
依题意，当13n 时，数列{}n a 是首项为20，公比为1
2
的等比数列，
所以13120522n n n a --⎛⎫
=⨯=⨯ ⎪⎝⎭
．故210a =，35a =，所以321a a a <<．
·············································································································· 2分 当4n 时，数列{}n a 是以4a 为首项，公差为4的等差数列，又4349a a =+=，所以()94447n a n n =+-⨯=-． ········································································· 3分
令20n a >，得27
4
n >
，又因为*n ∈N ，所以7n ． ·········································· 4分 因此，第7年初M 开始的价值n a 超过原购买的价值． ······································· 5分
（Ⅱ）设n S 表示前n 年初M 的价值的和，则n n S
T n
=
由（Ⅰ）知，当13n 时，3120124052112
n n
n S -⎛⎫⎛⎫⋅- ⎪
⎪ ⎪⎝⎭⎝⎭==-⨯-，34052n n T n --⨯=①； ·············································································································· 7分 当4n 时，由于335S =，故
()()()23453947 (3525322)
n n n n S S a a a n n -+-=++++=+
=-+，
2
253232
25n n n T n n n
-+==+-．② ·
······························································· 9分 当13n 时，由①得，120T =，215T =，335
3
T =，所以123T T T >> ················ 10分
当4n 时，由②知，32322522511n T n n n n =+-⋅=，当且仅当322n n
=，即4n =时等号成立．即()4min 11n T T ==． ··································································· 11分 由于34T T >，故在第4年初n T 的值最小，其最小值为11． ································ 12分 21．本小题主要考查函数、导数、不等式等基础知识，考查推理论证能力、抽象概括能力、运算求解能力，考查函数与方程思想、数形结合思想、化归与转化思想．满分12分．
解：（Ⅰ）因为()e x x
f x m
=+，所以22
e e e (1)()(e )(e )x x x x x m x x m
f x m m +-⋅-+'==++， ············· 1分 由1x =是()f x 的极值点，得2
(1)0(e )m
f m '==+， ·
············································ 2分 解得0m =， ····························································································· 3分
此时()e
x x
f x =，经检验，1x =是()f x 的极值点．
所以所求的实数m 的值为0． ······································································· 4分
（Ⅱ）证明：取1m =-时，()e 1
x x
f x =-，此时2e (1)1()(e 1)x x x f x --'=-． ···················· 6分
构造函数()e (1)1x h x x =--， ········································································ 7分 所以()e (1)e (1)e x x x h x x x '=-+-=-在(0,)+∞上恒负，
所以()h x 在(0,)+∞上单调递减， ··································································· 8分 所以()(0)0h x h <=, ···················································································· 9分
故()0f x '<在(0,)+∞恒成立，说明()e 1
x x
f x =-在(0,)+∞上单调递减． ··············· 10分
所以当01a b <<<时，e 1e 1
b a
b a
<--，又因为e e 1b a >>，所以e 10,e 10b a ->->， 所以(e 1)(e 1)a b
b a -<-， ··········································································· 11分 所以e e a b b a a b +<+成立． ········································································ 12分 22．本小题主要考查直线、椭圆等基础知识，考查推理论证能力、运算求解能力，考查函数与方程思想、数形结合思想、化归与转化思想、分类与整合思想．满分14分． 【解析】（Ⅰ）设椭圆Γ的半焦距为
c ，则
1,
23,c e a AF a c ⎧
==⎪⎨
⎪=+=⎩ ·································································· 2分 解得2,1a c ==， ······················································································· 3分 所以2223b a c =-=， ················································································· 4分
所以椭圆Γ的方程为22
1
x y +=． ·
································································ 5分
设l 方程为()1y k x =+（0k ≠），()()1122,,,P x y Q x y ． 因为点Q 与点Q '关于x 轴对称,所以()22,Q x y '-．
又因为椭圆关于x 轴对称，所以点Q '也在椭圆Γ上． ········································ 6分 由()22
1,3412,
y k x x y ⎧=+⎨+=⎩消去y 得()
22223484120k x k x k +++-=． 所以22121222
8412
0,,3434k k x x x x k k
-∆>+=-⋅=++． ················································· 7分 因为PF AQ '∥，所以直线AQ '的方程为()2y k x =-．
由()22
2,3412,
y k x x y ⎧=-⎨+=⎩消去y 得()
2222341616120k x k x k +-+-=． 因为直线AQ '交椭圆于()()222,0,,A Q x y '-两点，
所以222
1612234k x k -⋅=+，故22286
34k x k -=+． ························································· 9分
所以22221211212222
86886412
0,,34343434k k k k x x x x x x k k k k
---∆>+=+=-⋅=⋅=++++， 解得2157
,44k x ==-．
所以222861
342
k x k -==+． ·
············································································· 11分 所以()()()()
()22222
22211111811111164PF x y x k x k x =++=+++=++=， ················ 12分
()()()()
()222222
2222222812221216
AQ x y x k x k x '=-+=-+-=+-=． ·················· 13分 所以1
2
PF AQ '=． ·················································································· 14分
方法三：依题意，得PQ 与坐标轴不垂直．
设l 方程为()1y k x =+（0k ≠），()()1122,,,P x y Q x y ． 因为点Q 与点Q '关于x 轴对称,所以()22,Q x y '-．。