福建省福州市2014届高三上学期期末质量检测数学理试题扫描版含答案

数学(理科)试卷参考答案与评分标准
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答案中有且只有一个答案是正确的．把正确选项涂在答题卡的相应位置上．）
1. C
2. B
3. B 4．A 5. B 6. A 7. D 8. B 9. C 10．C 11. B 12. B
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．）
13．1 14． 15．22
2
n n -+ 16.．②③④
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）
解: (Ⅰ)x b x g 2sin 1)(2
2
=-=→-
··········································· 2分
由0)(=x g 得()Z k k x x ∈=∴=π202sin 即 ()Z k k x ∈=2π
····························· 5分 故方程)(x g ＝0的解集为{
()}Z k k x x ∈=
2
π
······················································· 6分 (Ⅱ)12sin 3cos 21)2sin ,1()3,cos 2(1)(2
2
-+=-⋅=-⋅=→
-→-x x x x b a x f ······ 7分 )6
2sin(22sin 32cos π
+=+=x x x ·
··················································· 9分 ∴函数)(x f 的最小周期ππ
==2
2T ·
······································································ 10分 由()Z k k x k ∈+≤
+
≤+-
ππ
π
ππ
22
6
222
得()Z k k x k ∈+≤
≤+-
ππ
ππ
6
3
故函数)(x f 的单调增区间为()Z k k k ∈⎥⎦
⎤
+⎢⎣⎡+-
ππππ6,3. （ 开区间也可以）
··································································································································· 12分
18. （本小题满分12分） 解:(Ⅰ)
1111
,033n n n n a a a a n ++==∴>
1111==n 13n 13
n n
a a a +∴+，又 ·
······································································· 2分 n n a ⎧⎫
∴⎨⎬⎩⎭
11为首项为，公比为的等比数列33 ·
············································· 4分
n 1
n
11n
==
n 333n n a a -⎛⎫∴⨯∴ ⎪⎝⎭
， ··············································································· 6分 (Ⅱ) 1231233333n n
n
S =
++++……① ····································································· 7分 231112133333
n n n n n
S +-∴=++++……② ·
·················································· 8分 ①-② 得：123121111333333
n n n n
S +=++++- ··································· 9分
1
11
13313
13
n n n +⎛⎫-
⎪
⎝⎭=
-- ·
·················································· 10分 31
143
23
n n
n
n S ⎛⎫
∴=-- ⎪⨯⎝⎭ 133243
n n n
n
S +--∴=⨯ ················································································· 12分
19. （本小题满分12分） .解：(Ⅰ)根据题意，
分别记“甲所付租车费0元、1元、2元”为事件123,,A A A ，它们彼此互斥， 且123()0.4,()0.5,()10.40.50.1P A P A P A ==∴=--=
分别记“乙所付租车费0元、1元、2元”为事件123,,B B B ，它们彼此互斥， 且123()0.5,()0.3,()10.50.30.2P B P B P B ==∴=--= ····················· 2分 由题知，123,,A A A 与123,,B B B 相互独立， ········································· ３分 记甲、乙两人所扣积分相同为事件M ，则112233M A B A B A B =++ 所以112233()()()()()()()P M P A P B P A P B P A P B =++
0.40.50.50.30.10.20.20.150.020.37=⨯+⨯+⨯=++= ······ ６分 (Ⅱ) 据题意ξ的可能取值为：0,1,2,3,4 ·········································· 7分 11(0)()()0.2P P A P B ξ===
1221(1)()()()()0.40.30.50.50.37P P A P B P A P B ξ==+=⨯+⨯=
132231(2)()()()()()()0.40.20.50.30.10.50.28
P P A P B P A P B P A P B ξ==++=⨯+⨯+⨯= 2332(3)()()()()0.50.20.10.30.13P P A P B P A P B ξ==+=⨯+⨯= 33(4)()()0.10.20.02P P A P B ξ===⨯= ············································· 10分
的数学期望 ···· 11分 答：甲、乙两人所扣积分相同的概率为0.37，ξ的数学期望 1.4E ξ= ··············· 12分
20．（本小题满分12分）
解：依题意得g(x)3x =+，设利润函数为f(x)，则f(x)(x)g(x)r =-，
所以20.5613.5(0x 7)f(x),10.5(x 7)
x x x
⎧-+-≤≤=⎨
->⎩ ································· 2分
（I ）要使工厂有盈利，则有f (x )＞0，因为
f (x )＞0⇔2
0x 77
0.5613.5010.50x x x x ≤≤>⎧⎧⎨⎨-+->->⎩⎩
或， ···························· 4分 ⇒20x 771227010.50
x x x x ≤≤>⎧⎧⎨⎨
-+<->⎩⎩或⇒0x 7710.5
39x x ≤≤⎧<<⎨<<⎩或
⇒3x 7<≤或7x 10.5<， ················································ 6分
即3x
10.5<. ···································································· 7分
所以要使工厂盈利，产品数量应控制在大于300台小于1050台的范围内． ···· 8分 （II ）当3x 7<≤时， 2
f(x)0.5(6) 4.5x =--+
故当x ＝6时，f (x )有最大值4.5. ···················································· 10分 而当x ＞7时，f(x)10.57 3.5<-=.
所以当工厂生产600台产品时，盈利最大． ········································· 12分
21. （本小题满分12分）
s 解：（I ）设双曲线C 的方程为22
221(00)x y a b a b
-=>>，， ····························· 1分
由题设得229a b b a ⎧+=⎪
⎨=⎪⎩，
·················································································· 2分
解得2
245.a b ⎧=⎪⎨=⎪⎩，，
····································································································· 3分
所以双曲线C 的方程为22
145
x y -=； ····························································· 4分 （II ）设直线l 的方程为(0)y kx m k =+≠，点11()M x y ，，22()N x y ，的坐标满
足方程组2
2
1.45y kx m x y =+⎧⎪⎨-
=⎪⎩， ① ②，
将①式代入②式，得22
()145x kx m +-=，
整理得222
(54)84200k x kmx m ----=， ·················································· 6分 此方程有两个不等实根，于是2540k -≠， 且2
2
2
(8)4(54)(420)0km k m ∆=-+-+>，
整理得22540m k +->.③ ··········································································· 7分 由根与系数的关系可知线段MN 的中点坐标00()x y ，满足：
12024254x x km x k +=
=-，002
554m
y kx m k
=+=-， ································ 8分 从而线段MN 的垂直平分线的方程为225145454m km y x k k k ⎛⎫
-
=-- ⎪--⎝⎭，
···· 9分 此直线与x 轴，y 轴的交点坐标分别为29054km k ⎛⎫
⎪-⎝⎭，，29054m k ⎛
⎫ ⎪
-⎝⎭
，， 由题设可得
2
2
199********
km
m k k =--，整理得222
(54)k m k -=，0k ≠， ································································································································· 10分
将上式代入③式得22
2(54)540k k k
-+->， ·
··········································· 11分
整理得2
2
(45)(45)0k k k --->，0k ≠，解得0k <<
或5
4
k >， 所以k 的取值范围是55550044⎛⎫⎛⎫⎛
⎫⎛⎫
---+ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭
⎝⎭⎝⎭
⎝∞，，，，∞． ······· 12分 22. （本小题满分14分）
解：（Ⅰ）当2a =时，2()ln(1)1
x
f x x x =+++， ∴22
123
()1(1)(1)x f x x x x +'=
+=+++， ······································································ 1分 ∴ (0)3f '=，所以所求的切线的斜率为3. ··························································· 2分 又∵()00f =，所以切点为()0,0. ····································································· 3分 故所求的切线方程为：3y x =. ··········································································· 4分 （Ⅱ）∵()ln(1)1
ax
f x x x =++
+(1)x >-，
∴22
1(1)1()1(1)(1)a x ax x a
f x x x x +-++'=
+=
+++． ··························································· ６分 ①当0a ≥时，∵1x >-，∴()0f x '>； ······························································ ７分 ②当0a <时，
由()01f x x '<⎧⎨>-⎩，得11x a -<<--；由()0
1
f x x '>⎧⎨>-⎩，得1x a >--； ····················· ８分 综上，当0a ≥时，函数()f x 在(1,)-+∞单调递增；
当0a <时，函数()f x 在(1,1)a ---单调递减，在(1,)a --+∞上单调递增． ····· ９分 （Ⅲ）方法一：由（Ⅱ）可知，当1a =-时， ()()ln 11
x
f x x x =+-
+在()0,+∞上单调递增． ·················································· 10分 ∴ 当0x >时，()()00f x f >=，即()ln 11
x
x x +>
+． ································· 11分 令1x n =
（*n ∈N ），则111ln 111
1n
n n n
⎛⎫
+>= ⎪+⎝⎭+． ············································· 12分
另一方面，∵
()211
1n n n
<+，即21111n n n -
<+, ∴
2111
1n n n
>-+．
······························································································ 13分 ∴ 2111
ln 1n n n
⎛⎫+>- ⎪⎝⎭（*n ∈N ）． ····································································· 14分
方法二：构造函数2
()ln(1)F x x x x =+-+，(01)x ≤≤ ································· 10分 ∴1(21)
'()1211
x x F x x x x +=
-+=
++， ······························································ 11分 ∴当01x <≤时,'()0F x >；
∴函数()F x 在(0,1]单调递增． ·········································································· 12分 ∴函数()(0)F x F > ，即()0F x >
∴(0,1]x ∀∈，2
ln(1)0x x x +-+>，即2
ln(1)x x x +>- ···························· 13分 令1x n =（*n ∈N ），则有2111ln 1n n n ⎛⎫+>- ⎪⎝⎭
． ·················································· 14分。