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【全国省级联考word】四川省2018届高三春季诊断性测试理数试题

【全国省级联考word】四川省2018届高三春季诊断性测试理数试题
【全国省级联考word】四川省2018届高三春季诊断性测试理数试题

数学(理科)

第Ⅰ卷

一、选择题:本大题共12个小题,每小题5分,共60分.在每小题给出的四个选项中,只有一项是符合题目要求的.

1.已知集合2{|4}A x x =>,集合{|31}B x x =-<<,则A B ?=( ) A .()3,2-- B .(),1-∞ C .()3,1- D .()(),12,-∞?+∞

2.若向量()21,m k k =-与向量()4,1n =共线,则m n =( ) A .0 B .4 C .92-

D .172

-

3.若虚部大于0的复数z 满足方程2

40z +=,则复数

1z

z

+的共轭复数为( ) A .4255i + B .4255i - C .4255i -+ D .4255

i --

4.某几何体的三视图如图所示,其中俯视图中的圆的半径为2,则该几何体的体积为( )

A .51296π-

B .296 C. 51224π- D .512

5.设,x y 满足约束条件330

280440x y x y x y -+≥??

+-≤??+-≥?

,则3z x y =+的最大值是( )

A .9

B .8 C. 3 D .4 6.若2

1sin 22cos

2

x

x +=,()0,x π∈,则tan 2x 的值构成的集合为( ) A

. B

.{

C.{ D

.{}33

- 7.执行如图所示的程序框图,则输出的S =( )

A .2

B . 1 C. 0 D .-1

8.9

3)x

的展开式中不含3

x 项的各项系数之和为( )

A .485

B .539 C.-485 D .-539

9.已知函数()f x 为偶函数,当0x >时,()4x f x -=,

设3(log 0.2)a f =,0.2(3)b f -=, 1.1(3)c f =-,则( )

A .c a b >>

B .a b c >> C. c b a >> D .b a c >>

10.过双曲线2

2

:13

y M x -

=的左焦点F 作圆221:(3)2C x y +-=的切线,此切线与M 的左支、右支分别交于A ,B 两点,则线段AB 的中点到x 轴的距离为( ) A . 2 B .3 C. 4 D .5

11.将函数sin 2y x x =的图象向左平移(0)2

π

??<≤个单位长度后得到()f x 的图象.若()

f x 在(

,)42ππ

上单调递减,则?的取值范围为( ) A .[,]32ππ B .[,]62ππ C. 5[,

]312ππ D .5[,]612

ππ

12.已知直线l 是曲线x

y e =与曲线2x

y e =-的一条公切线,l 与曲线22x

y e =-切于点(),a b ,且a 是函

数()f x 的零点,则()f x 的解析式可能为( )

A .()2(22ln21)1x

f x e x =+-- B .()2(22ln21)2x

f x e x =+--

C. ()2(22ln21)1x

f x e x =--- D .()2(22ln21)2x

f x e x =---

第Ⅱ卷

二、填空题(每题5分,满分20分,将答案填在答题纸上)

13.我国古代数学名著《九章算术》有一抽样问题:“今有北乡若干人,西乡七千四百八十八人,南乡六千九百一十二人,凡三乡,发役三百人,而北乡需遣一百零八人,问北乡人数几何?“其意思为:“今有某地北面若干人,西面有7488人,南面有6912人,这三面要征调300人,而北面共征调108人(用分层抽样的方法),则北面共有 人.”

14.若椭圆

22

14x y m

+=上一点到两个焦点的距离之和为3m -,则此椭圆的离心率为 .

15.在ABC ?中,sin B A =,BC =4

C π

=

,则AB 边上的高为 .

16.在底面是正方形的四棱锥P ABCD -中,PA ⊥底面ABCD ,点E 为棱PB 的中点,点F 在棱AD 上,平面CEF 与PA 交于点K , 且3PA AB ==,2AF =,则四棱锥K ABCD -的外接球的表面积为 .

三、解答题 (本大题共6小题,共70分.解答应写出文字说明、证明过程或演算步骤.)

17.设n S 为数列{}n a 的前n 项和,已知37a =,()12222n n a a a n -=+-≥. (1)证明:{1}n a +为等比数列; (2)求n S .

18. 根据以往的经验,某建筑工程施工期间的降水量N (单位:mm )对工期的影响如下表:

根据某气象站的资料,某调查小组抄录了该工程施工地某月前20天的降水量的数据,绘制得到降水量的折线图,如下图所示.

(1)根据降水量的折线图,分别求该工程施工延误天数0,1,3,6X =的频率; (2)以(1)中的频率作为概率,求工期延误天数X 的分布列及数学期望与方差.

19.如图,在直三棱柱111ABC A B C -中,12AC AA ==,D 为棱1CC 的中点,11AB A B O ?=

.

(1)证明:1//C O 平面ABD ; (2)设二面角D AB C --

AC BC ⊥,E 为线段1A B 上一点,且

CE 与平面ABD 所成

1

BE

BA . 20. 已知曲线M 由抛物线2x y =-及抛物线2

4x y =组成,直线:3(0)l y kx k =->与曲线M 有

()m m N ∈个公共点.

(1)若3m ≥,求k 的最小值;

(2)若4m =,自上而下记这4个交点分别为,,,A B C D ,求||

||

AB CD 的取值范围. 21. 已知函数()()()ln 1+ln 1f x x x =--.

(1)讨论函数()()()0F x f x ax a =+≠的单调性;

(2)若()3

(3)f x k x x >-对()0,1x ∈恒成立,求k 的取值范围.

请考生在22、23两题中任选一题作答,如果多做,则按所做的第一题记分.

22.选修4-4:坐标系与参数方程

在平面直角坐标系xOy 中,圆C 的参数方程为2(1cos )

2sin x y αα

=+??

=?,(α为参数),以坐标原点为极点,x 轴

的正半轴为极轴建立极坐标系.已知直线l 的极坐标方程为0θθ=,0(0,)2

π

θ∈,且0tan θ=

. (1)求圆C 的极坐标方程;

(2)设M 为直线l 与圆C 在第一象限的交点,求||OM . 23.选修4-5:不等式选讲 已知函数()1|2|f x x =--.

(1)求不等式()1|4|f x x >-+的解集;

(2)若()||f x x m >-对5(2,)2

x ∈恒成立,求m 的取值范围.

试卷答案

一、选择题

1-5:ADBCA 6-10: CBCAB 11、12:DB

二、填空题

16.48625π 三、解答题

17.(1)证明:∵37a =,3222a a =-,∴23a =,

∴121n n a a -=+,∴11a =, 则

()111122

2211

n n n n a a n a a ---++==≥++,

∴{1}n a +是首项为2,公比为2的等比数列. (2)解:由(1)知,12n n a +=,则21n n a =-. ∴2(222)n n S n =++

+-122n n +=--.

18.解:(1)∵400N mm <的天数为10,∴0X =的频率为10

0.520

=. ∵400600mm N mm ≤<的天数为6,∴1X =的频率为

6

0.320=. ∵6001000m N mm ≤<的天数为2,∴3X =的频率为2

0.120

=. (2)X 的分布列为

()00.510.330.1 1.2E X =?+?+?=.

()()2

0 1.20.5D X =-?+()()2

2

1 1.20.33 1.2-?+-()2

0.16 1.20.1?+-? 0.720.0120.324 2.304 3.36=+++=.

19.(1)证明:取AB 的中点F ,连接,OF DF ,

∵侧面11ABB A 为平行四边形,∴

O 为1AB 的中点, ∴11//

2OF BB ,又111

//2

C D BB ,∴1//OF C D , ∴四边形1OFDC 为平行四边形,则1//C O DF .

∵1C O ?平面ABD ,DF ?平面ABD ,∴1//C O 平面ABD . (2)解:过C 作CH AB ⊥于H ,连接DH , 则DHC ∠即为二面角D AB C --的平面角. ∵1DC =

,tan 2

DHC ∠=

,∴CH =.

又2AC =,AC BC ⊥,∴2BC =.

以C 为原点,建立空间直角坐标系C xyz -,如图所示,则()2,0,0A ,()0,2,0B ,()0,0,1D ,()2,0,2A , 则()2,2,0AB =-,()0,2,1BD =-,设平面ABD 的法向量(),,n x y z =, 则0AB n BD n ?=?=,即2220x y y z -+=-+=,令1y =,得()1,1,2n =.

设()101BM BA λλ=≤≤,∵()12,2,2BA =-,∴1CE CB BA λ=+()2,22,2λλλ=-, ∴CE 与平面ABD 所成角的正弦值为|cos ,|CE n <

>=|=

=

∴2

3644130λλ-+=,∴12λ=

或1318,即112BE BA =或1318

.

20.解:(1)联立2

x y =-与3y kx =-,得2

30x kx +-=,

∵21120k ?=+>,∴l 与抛物线2x y =-恒有两个交点.

联立2

4x y =与3y kx =-,得2

4120x kx -+=.

∵3m ≥,∴2216480k ?=-≥.

∵0k >

,∴k ≥k

(2)设11(,)A x y ,22(,)B x y ,33(,)C x y ,44(,)D x y ,

则,A B 两点在抛物线2

4x y =上,,C D 两点在抛物线2

x y =-上,

∴124x x k +=,1212x x =,34x x k +=-,343x x =-,且2

216480k ?=->,0k >

,∴k >

||AB =

||CD =

||||

AB CD =

=

=∴k >2

150112k <

<+,∴()||0,4||

AB CD ∈. 21.解:(1)()11'11F x a x x =+++-()22

2

111ax a x x -++=-<<-, 当20a -≤<时,()'0F x ≥,∴()F x 在()1,1-上单调递增.

当0a >时,()'0F x >,故当20a -≤<或0a >时,()F x 在()1,1-上单调递增. 当2a <-时,令(

)'0F x >,得1x

-<<1x <<; 令

()'0F x <

,得x

<

∴()F

x 在(

上单调递减,在(1,

-,上单调递增. (2)设()()()

3

3g x f x k x x =--,则()()

2

22

231'1k x

g x x

+-=

-,

当()0,1x ∈时,()()2

210,1x -∈,或2

3

k ≥-,()22310k x +->,则()'0g x >, ∴()g x 在()0,1上递增,从而()()00g x g >=.

此时,()()

3

3f x k x x >-在()0,1上恒成立.

若23k <-

,令()'0g x x

=?=()0,1

,当x ∈时,()'0g x <;

当x ∈时,()'0g x >. ∴

()()min 00g x g g =<=,则23k <-不合题意.

故k 的取值范围为2

[,)3

-+∞.

22.解:(1)由()21cos 2sin x y αα

=+???=??,消去α得()22

24x y -+=,

∴224x y x +=,∴24cos ρρθ=,

即4cos ρθ=,故圆C 的极坐标方程为4cos ρθ=.

(2)∵0(0,)2

π

θ∈,且0tan 3

θ=

,∴03cos 4θ=.

将3

cos 4

θ=

代入4cos ρθ=,得3ρ=, ∴||3OM =.

23.解:(1)由()1|4|f x x >-+,得|2||4|x x -<+,

不等式两边同时平方得2

2

44816x x x x -+<++,解得1x >-, ∴所求不等式的解集为()1,-+∞.

(2)当5(2,)2

x ∈时,()()123f x x x =--=-. ∴3||x x m ->-,即33

x m x x m x -<-??

->-?,对5

(2,)2x ∈恒成立,

即323

m x m +>??

(2,)2x ∈恒成立,又()24,5x ∈,∴35m +≥且3m <,

∴[2,3)m ∈

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