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
福州市2013—2014学年第一学期高三期末质量检测
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)
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
1111331313
n n n
+??
- ???=-- ·
··············································· 10分 3114323n 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()()()()()()()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 ≤≤?<<
?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)
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2
2
2
(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 +=
=-,00
2
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22
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 ≠,解得或5
4
k >,
所以k 的取值范围是550044???????
--+ ? ? ? ? ???
???? ∞,,∞. ······ 12分 22. (本小题满分14分)
解:(Ⅰ)当2a =时,2()ln(1)1
x
f x x x =+++,ks5u ∴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
1f 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1
11ln 1111n n n n ??
+>= ?+??+. ··········································· 12分
另一方面,∵
()2111n n n <+,即2111
1n 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 ?∈,2ln(1)0x x x +-+>,即2ln(1)x x x +>- ··························· 13分 令1x n =(*n ∈N ),则有2111ln 1n n n
??+>- ???. ················································ 14分