四川省泸州市2015届高三第二次教学质量诊断性考试数学(理)试题 Word版含答案

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泸州市高2012级第二次教学质量检测

数 学（理工类）

本试卷分第一部分（选择题）和第二部分（非选择题）两部分. 第一部分1至2页，第二部分3至4页.共150分.考试时间120分钟. 第一部分的答案涂在机读卡上，第二部分的答案写在答题卡上.

第一部分（选择题 共50分）

注意事项：

1. 答第一部分前，考生务必将自己的姓名、准考证号、考试科目用铅笔涂、写在机读卡上.

2. 每小题选出答案后，用2B 铅笔把机题卡上对应题的答案标号涂黑，如需改动，用橡皮擦干净后，再选涂其它答案，不能答在草稿子、试题卷上.

3．本部分共12个题，每小题5分共60分.

一、选择题：本大题共10小题，每小题5分，共50分.在每小题给出的四个选项中只有一个选 项是符合题目要求.

1．已知集合{|12}A x x =-<<，2{|log 2}B x x =<，则A

B =

A ．(1,2)

B ．(1,4)-

C ．(0,2)

D ．(0,4)

2．计算sin 43cos13sin13sin 47-的值等于

A ．

12

B

C

．

D

3．函数ln ||

||

x x y x =

的图象可能是

4．下列命题中，真命题是

A ．0x ?∈R ，020x ≤

B ．x ?∈R ，22x x >

C ．“0a b +=”是“

1a

b

=-”的充要条件 D ．“1,1a b >>”是“1ab >”的充分条件 5． 执行如图所示的程序框图，输出S 的值是

A ．3

B ．－6

C ．10

D ．－

15

- 2 -

6． 一个几何体的三视图如图所示，则该几何体的表面积是

A ．5π

B ．6π

C ．7π

D ．8π 7． 不等式组1

24x y x y +??-?

≥≤的解集记为D ，下列四个命题中正确的是

A ．(,)x y D ?∈，22x y +-≥

B ．(,)x y D ?∈，22x y +≥

C ．(,)x y

D ?∈，23x y +≤ D ．(,)x y D ?∈，21x y +-≤

8. 某工厂某种产品的年固定成本为250万元，每生产x 千件需另投入成本为()G x ，当年产量不足80千件时，21()103G x x x =+(万元)；当年产量不小于80千件时，10000

()511450G x x x

=+-(万元)．

每件商品售价为0.05万元．通过市场分析，该厂生产的商品能全部售完，则该厂在这一商品的生产

中所获年利润的最大值是 A ．900万元

B ．950万元

C ．1000万元

D ．1150万元

9． 已知一个三角形的三边长分别是5，5，6，一只蚂蚁在其内部爬行，若不考虑蚂蚁的大小，则某时刻

该蚂蚁距离三角形的三个顶点的距离均超过2的概率是 A ．2－

3

π

B ．1－

6

π

C ．2－

2

π

D ．1－

12

π

10．已知函数322

(1),()11(1)22,0.3

2kx k a x f x x ax a x a a x +-??

=?-+--+-

B ．2-

C ．4-

D ．3-

第二部分 （非选择题 共100分）

注意事项:

1．用0.5毫米黑色签字笔直接答在答题卡上，答在试题卷上无效.作图题可先用铅笔绘出，确认后再用0.5毫米黑色签字笔描清楚，答在试题卷上无效.

2．本卷共11个小题，共100分.

二、填空题：本大题共5小题，每小题5分，共25分，把正确答案填在题中横线上 11．如果复数2

1i

z =

-+（i 是虚数单位），则复数z 的虚部为 ． 4

2

侧视图

俯视图

正视图

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12．已知数列{}n a 为等差数列, n S 为其前n 项和，若11

2

a =，23S a =，则其公差为_____．

13．在三棱柱ABC -A 1B 1C 1中，∠BCA ＝90°，AA 1⊥平面ABC ，M ，N 分别是

A 1

B 1，A 1

C 1的中点，若BC ＝CA ＝CC 1，则直线BM 与直线AN 所成角的 余弦值为 .

14．已知ABC △的内角A ,B ,C 所对的边分别为a ，b ，c ，重心为G （三角形中

三边中线的交点），若233aGA bGB cGC ++=0，则cos B = ．

16．（本题满分12分）

以下茎叶图记录了甲、乙两组各5名学生在一次英语听力测试中的成绩(单位:分)：

已知甲组数据的中位数为13，乙组数据的众数是18.

（Ⅰ）求,x y 的值，并用统计知识分析两组学生成绩的优劣；

（Ⅱ）从两组学生中任意抽取3名，记抽到甲组的学生人数为X ，求X 的分布列和期望.

M

B 1

A 1

C 1

A

C

B

N

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17．（本题满分12分）

在锐角ABC △中，角A,B,C 所对的边分别为c b a ,,，向量(2,1)a m =，(cos ,2)C c b -n =，

且

⊥m n ．

（Ⅰ） 求角A 的大小；

（Ⅱ）求函数2cos2()11tan C

f C C

=-

+的值域．

18．（本题满分12分）

已知函数1()lg

(0)1ax f x a x +=>-为奇函数，函数22

()()g x b b x

=+∈R ． （Ⅰ）求函数()f x 的定义域;

（Ⅱ）当11

[,]32

x ∈时，关于x 的不等式(1)lg ()f x g x -≤有解，求b 的取值范围．

19．（本题满分12分）

已知数列{}n a ，满足1

1

,21,.

n

n n a n a a n +??=??+?为偶数，

为奇数，452a =，若211(0)n n n b a b -=-≠． （Ⅰ）求1a ，并证明数列{}n b 是等比数列；

（Ⅱ）令21(21)n n c n a -=-，求数列{}n c 的前n 项和n T ．

20．（本题满分13分）

如图，在多面体111-ABC A B C 中，侧面11AA B B ⊥底面111A B C ，四边形11AA B B 是矩形，1111AC =A B ，111120B AC ∠=，11//BC B C ，112B C BC =．

（Ⅰ）求证：1

11AC B C ⊥； （Ⅱ）当二面角11C AC B --的正切值为2时，求

1

11

AA A B 的值．

21．（本题满分14分）

已知函数3()3f x x x =-,()e ()x g x ax a =-∈R ．其中e 是自然对数的底数. （Ⅰ）求经过点2

(,2)3

A -与曲线()f x 相切的直线方程；

（Ⅱ）若函数()()1ln ((0,2])F x g x x x x =--∈，求证：当e 1a <-时，函数()F x 无零点；

（Ⅲ）已知正数m 满足：存在0[1)x ∈+∞，，使得000()()()g x g x mf x +-<-成立．试比较1e 1

e m m --与的大小，并证明你的结论．

C

A

A 1

C 1

B 1

B

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一、选择题

二、填空题

11．－1； 12．

12； 13； 14．112； 15. (1)(4) . 三、解答题

16．解：（Ⅰ）甲组五名学生的成绩为9,12,10+x ,24,27. 乙组五名学生的成绩为9，15，10+y ，18，24.

因为甲组数据的中位数为13，乙组数据的众数是18， 所以3x =， ·································································································································· 2分 8y =, ············································································································································· 4分 因为甲组数据的平均数为85

5， ·

································································································ 5分 乙组数据的平均数是84

5

， ········································································································· 6分

则甲组学生成绩稍好些； ··········································································································· 7分 （Ⅱ）X 的取值为0、1、2、3.

353101

(0)12C P X C ===， ·················································································································· 8分

1255310

5

(1)12C C P X C ===， ··············································································································· 9分

5

(2)12P X ==

， ························································································································· 10分 1

(3)12

P X ==， ························································································································· 11分 所以X EX =01123,121212122

?

+?+?+?= ∴X 的期望为3

.2

·························································································································· 12分

17．解：(Ⅰ) ∵⊥m n ， ∴2cos 20a C c b +-=， ····················································································································· 1分 由正弦定理得：2sin cos sin 2sin 0A C C B +-=， ·········································································· 2分 ∵()B A C π=-+， ···························································································································· 3分 ∴2sin cos sin 2sin()0A C C A C +-+=， ························································································· 5分 ∴2sin cos sin 2sin cos 2cos sin 0A C C A C A C +--=， ∴sin 2cos sin 0C A C -=，∵0C π<<，∴sin 0C >，

∴1cos 2A =，∴3

A π

=； ·················································································································· 6分

（Ⅱ）2cos 2()11tan C

f C C

-=++，

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222(cos sin )1sin 1cos C C C C

--=++

， ·················································································································· 7分

222cos (cos sin )

1cos sin C C C C C

--=++， ·

········································································································ 8分 22cos 2cos sin 1C C C =-++， ··········································································································· 9分 sin 2cos 2C C =-， ··························································································································· 10分

)4

C π

=- ·

······························································································································ 11分 ∵3A π=，ABC △是锐角三角形，∴62

C ππ

<<，∴321244C πππ<-<，

∴函数()f C

的值域为． ·

······················································································ 12分 18． 解：(Ⅰ)由1()lg (0)1ax

f x a x

+=>-为奇函数得：()()0f x f x -+=，

即22

2111lg lg lg 0111ax ax a x x x x -+-+==+--, ·

·························································································· 2分 所以22

2

111a x x -=-，解得1a =， ·

·································································································· 4分 经检验符合题意，故1()lg 1x

f x x

+=-, ··························································································· 5分

所以()f x 的定义域是(1,1)-； ···································································································· 6分

（Ⅱ）不等式(1)lg ()f x g x -≤等价于222

x b x x

-+≤， ·

························································· 7分 即222

1

b x

x -

+-≥在11[,]32x ∈有解， ························································································· 8分 故只需min 222

(1)b x x

-+-≥， ······································································································· 9分

因为11[,]32x ∈，所以1

[2,3]x ∈， ······························································································ 10分

函数2111

2()22

y x =---， ·········································································································· 11分

所以2min 11

2(3)1322

y =---=-，

所以13b -≥，所以b 的取值范围是[13,)-+∞. ······································································· 12分 19．解：（Ⅰ）∵452a =，11

,21,.

n n n a n a a n +??=??+?为偶数，

为奇数∴353122a =-=， ··································· 1分

∴23a =， ···································································································································· 2分

∴12a =； ····································································································································· 3分 ∵212+12-121

1

11

21112

n n n n n n a b a b a a +--===---， ······························································································ 5分 故数列{}n b 是首项为1，公比为1

2

的等比数列； ····································································· 6分

（Ⅱ）由（I ）知：11()2n n b -=，且1211

(21)(21)()(21)2

n n n c n a n n --=-=-+-， ························ 7分

- 7 -

1321

122

n n n S --=+++， ①

211132321

22222

n n n n n S ---=++++，②·

························································································ 8分 ①-②得：22111121

1122222n n n n S --=+++++-， ···································································· 9分

1

11()21211212

n n n ---=+--2332n n +=-， ·

························································································· 10分 所以46

62

n n n S +=-． ················································································································· 11分

故[135(21)]n n T S n =+++++-

246

62n

n n +=+-

. ························································································································· 12分 20．解法一：(Ⅰ) 取11B C 的中点D ，连接CD 、1A D ，因为11//BC B C ，112B C BC =， 所以1//CB DB ，∴1CB DB =， ∴四边形1CDB B 是平行四边形， ······································································································ 1分 又11AA B B 是矩形，∴1//CD AA ， ··········································· 2分 因为侧面11AA B B ⊥底面111A B C ，111AA A B ⊥， ∴1AA ⊥底面111A B C ，∴1AA ⊥11B C ， ····························· 3分 因为点D 是11B C 的中点，1111AC =A B ，∴111B C A D ⊥，

····· 4分 又111A D AA A =，∴11B C ⊥平面1AA DC ，∴1

11AC B C ⊥； ····················································· 6分 （Ⅱ）过点C 作CE AD ⊥于点E , 过点E 作1EF AC ⊥于点F, 连接CF ， ·································· 7分 由(Ⅰ)知：11B C ⊥平面1AA DC ，11B C ?平面11AB C , ∴平面11AB C ⊥平面1AA DC ，且交线为AD ，

CE ⊥平面11AB C ，即1CE AC ⊥， ·

··································································································· 8分 所以1AC ⊥平面CEF ， 则1CF AC ⊥

所以CFE ∠为二面角11C AC B --的平面角， ··········································································

······· 9分 设111A B =，1AA λ=

，则1C D ，1

12A D =，

所以AD =，CE =， ············································································

············· 10分

∵1AFE ADC △△，又因AE =1AC ∴22

14EF λ=

+, ··············································································································· 11分

二面角11C AC B --的正切值为2, ∴tan 2CE

CFE EF ∠==,∴

22

14λ=+, ···········································

···················

12分 解之得：λ=

111

AA A B ····························································

解法二：(Ⅰ) 过点1A 在平面111A B C 内作111EA A B ⊥,

因为侧面11AA B B ⊥底面111A B C ， 所以1EA ⊥侧面11AA B B ，

所以11A A A E ⊥，111A E A B ⊥， ··············································· 2分

C

- 8 -

因为11AA B B 是矩形，所以111A A A B ⊥，··························································································· 3分 以1A 为原点，射线1A E ，11A B , 1A A 的方向分别为x 轴，y 轴，z 轴的正方向，建立空间直角坐标系．

设111A B =，1AA λ=

，则1

C D ，112A D =，

所以1(0,1,0)B

，(0,0,)A λ，11

,0)2

C -，1

,)4C λ． ························································· 4分

因为131(,)AC λ=,11

33

(,0)2B C =-， 又111

313

()0042

AC B C λ=?-+?=， ·················································································· 5分 所以1

11AC B C ⊥，即111AC B C ⊥; ·························································

······················

························ 6分 （Ⅱ） 设平面11AB C 的一个法向量为(,,)x y z =m ，

所以1C A ⊥m ，11B C ⊥m ，

因为1(0,1,)B A λ=-，1133

(

,0)2

B C =-， 则00x y z λ?=??-=??

，取1)λ=m ， ·······

············································

······

···························

·····

········ 8分

设平面1AC C 的一个法向量为(,,)x y z =n ，所以1C A ⊥n ，CA ⊥n ， 因为11(,)2C A λ=-，31

(,0)4

AC =，

则0

20

y y z λ+=--=

，取(1,

=n ， ··································································

·······

·······

·· 10分

∵二面角11C AC B --的正切值为2,

∴二面角11C AC B --的余弦值为

, 即cos )<>=m,n ·················································· 11分

= ·

································································································· 12分 解得：λ=

111

AA A B ·············································································· 13分

21． 解：（Ⅰ）由3()3f x x x =-得2()33f x x '=-， 设过点A 的直线与曲线()f x 相切于点(,)B m n ，

所以切线斜率为2()33f m m '=-，

切线方程为323(33)()y m m m x m -+=--， ············································································ 1分

因为点2(,2)3A -在切线上，所以322

23(33)()3

m m m m -+=---， ········································ 2分

解之得：0m =，或1m =-

当0m =时，(0)3f '=-且(0,0)B ，故直线方程为3y x =-， ················································· 3分 当1m =-时，(1)0f '-=且(1,2)B -，故直线方程为2y =; ···················································· 4分

（Ⅱ）因为()e 1ln x F x ax x x =---，

由()0F x =得，e 1ln x a x x

-=-， ·

··································································································· 5分

- 9 -

设e 1()ln x h x x x -=-，则2

(e 1)(1)

()x x h x x --'=

， ············································································· 6分 当01x <<时，()0h x '<，当12x <<时，()0h x '> 所以()h x 在(0,1)单调递减，(1,2)单调递增， ················································································· 7分 又(1)e 1h =-，

所以当e 1a <-时，函数()F x 无零点；··························································································· 8分 （Ⅲ）()()()e +e x x G x g x g x -=+-=，则'()e e x x G x -=-，当1x >时'()0G x >，

∴()G x 在(1)+∞，上单调递增， ································································································· 9分

令3()()(3)h x mf x m x x =-=-+，2'()3(1)h x m x =--，

∵01m x >>，，∴'()0h x <，即()h x 在(1)x ∈+∞，上单调递减， ········································ 10分

∵存在0[1)x ∈+∞，

，使得30000()()(3)g x g x m x x +-<-+， ∴1(1)e 2e G m =+<，即11e 2e m ??

>+ ???

， ·················································································· 11分

∵e-1

e 111ln ln ln e (e 1)ln 1e

m m m m m m ---=-=--+， ········································································ 12分

设()(e 1)ln 1H m m m =--+，

则e 1e 111'()1e 2e m H m m m m ---??

=

-=>+ ???

，， ·

········································································· 13分 当11e e 12e m ??

+<<- ???

时，'()0H m >，()H m 单调递增， 当e 1m >-时，'()0H m <，()H m 单调递减，

因此()H m 至多有两个零点，而(1)(e)0H H ==，

∴当e m >时，()0H m <，e 11e m m --<；

当11e e 2e m ??

+<< ???

时，()0H m >，e 11>e m m --； 当e m =时，()0H m =，e 11=e m m --． ····················································································· 14分