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数学(强化班)卷·2019届江苏省天一中学高一上学期期末考试(2017.01)

数学(强化班)卷·2019届江苏省天一中学高一上学期期末考试(2017.01)
数学(强化班)卷·2019届江苏省天一中学高一上学期期末考试(2017.01)

江苏省天一中学2016年秋学期期末考试

高一强化班数学学科

命题人:王义 审核:李维维

注意事项及答题要求:

1.本试卷包含填空题(第1题~第14题,共14题)和解答题(第15题~第20题,共6 题)两部分.本次考试时间为120分钟,满分为160分.考试结束后,请将答题纸交回. 2.答题前,请务必将自己的班级、姓名、学号用黑色笔写在答题纸上密封线内的相应位置. 3.答题时请用黑色笔在答题纸上作答.......

,在试卷或草稿纸上作答一律无效. 一、填空题:本大题共14小题,每小题5分,共70分.请把答案直接填写在答题..纸相应位....置上..

. 1、已知{x 22},{1}M x N x

x =-≤≤=<,则()R C M N ?=___▲_____.

2、设,x y R ∈,向量(,1)a x =,(1,)b y =,(2,4)c =-,且,//a c b c ⊥,则x y + = ▲ _ .

3、已知向量,a b 夹角为45°,且||1,|2|10a a b =-=,则||b =___▲___.

4、若cos α=

且(,0)2πα∈-,则()sin πα-= ▲ .

5、设25a b m ==,且11

2a b +=,则m = ▲ . 6、将函数sin(2)3

y x π=-的图象先向左平移

3

π

个单位,再将图象上各点的横坐标变为原来

1

2

倍(纵坐标不变),那么所得到图象的解析式为y = ▲ . 7、若函数1x

1y ()

m 2

-=+的图象与x 轴有公共点,则m 的取值范围是__▲______.

8、设向量,a b 满足|a |=(2,1)b =,且a 与b 的方向相反,则a 的坐标为___▲____. 9、若θ是△ABC 的一个内角,且sinθcosθ=-错误!未指定书签。,则sinθ-cosθ的值为 ▲ . 10、已知角φ的终边经过点P(1,-2),函数f(x)=sin(ωx+φ)(ω>0)图象的相邻两条对称轴之间的距离等于错误!未指定书签。,则f(错误!未指定书签。)= ▲ .

11、已知f (x )=?

??

??

3-a x -a x <

log a x x 是(-∞,+∞)上的增函数,那么实数a 的取

值范围是____▲____.

12、如图,在ABC ?中,D 是BC 的中点,,E F 是,A D 上的两个三等分点,

4BA CA ?=,1BF CF ?=- ,则BE CE ? 的值是 ▲ .

13、对于实数a 和b ,定义运算“*”:?????>-≤-=*b

a a

b b b

a a

b a b a ,,22

, 设

)1()12()(-*-=x x x f ,且关于x 的方程为f (x )=m (m ∈R )恰有三个互不相等的实数

根123x x x 、、,则123x x x ++的取值范围是_____▲________. 14、已知函数()sin()(0),2

4

f x x+x π

π

ω?ω?=>≤

=-

, 为()f x 的零点,4

x π

=

()y f x =图像的对称轴,且()f x 在51836ππ??

???

,单调,则ω的最大值为 ▲ .

二、解答题:本大题共6题,共90分,解答应写出文字说明、证明过程或演算步骤. (见答题纸)

15、(本题14

分)设函数2()2cos f x x x ωω=

+,其中02ω<<. (Ⅰ)若()f x 的最小正周期为π,求()f x 的单调增区间; (Ⅱ)若函数()f x 的图象的一条对称轴为3

x π

=,求ω的值.

16、(本题14分)已知△ABC 中.

(1)设→BC ?→CA =→CA ?→

AB ,求证:△ABC 是等腰三角形;

(2)设向量s →=(2sinC ,-3),t →=(sin 2C ,2cos 2c 2-1),且s →//t →

,若sinA =13,求si n (π3-B )

的值.

17、(本题14分)如图,半径为1,圆心角为3π

2的圆弧AB 上有一点C .

(1)若C 为圆弧AB 的中点,点D 在线段OA 上运动,求|→OC +→

OD |的最小值;

(2)若D ,E 分别为线段OA ,OB 的中点,当C 在圆弧AB 上运动时,求→CE ?→

CD 的取值范围.

18、(本题16分)某仓库为了保持库内的湿度和温度,四周墙上均装有如图所示的自动通风设施.该设施的下部ABCD 是矩形,其中AB =2米,BC =0.5米,上部CmD 是个半圆,固定点E 为CD 的中点.△EMN 是由电脑控制其形状变化的三角通风窗(阴影部分均不通风),MN 是可以沿设施边框上下滑动且始终保持和AB 平行的伸缩横杆(MN 和AB ,DC 不重合).

(1)当MN 和AB 之间的距离为1米时,求此时三角通风窗EMN 的通风面积;

(2)设MN 与AB 之间的距离为x 米,试将三角通风窗EMN 的通风面积S(平方米)表示成关于x 的函数S =f(x);

(3)当MN 与AB 之间的距离为多少米时,三角通风窗EMN 的通风面积最大?并求出这个最大面积.

P

19、(本题16分)如图,正方形ABCD 中边长为1,P 、Q 分别为BC 、CD 上的点,CPQ ?周长为2.

(1)求PQ 的最小值;

(2)试探究求PAQ ∠是否为定值,若是给出证明;不是说明理由.

20、(本题16分)已知函数()2f x x x a x =-+.

(1)若函数()f x 在R 上是增函数,求实数a 的取值范围;

(2)求所有的实数a ,使得对任意[1,2]x ∈时,函数()f x 的图象恒在函数()21g x x =+图 象的下方;

(3)若存在[4,4]a ∈-,使得关于x 的方程()()f x t f a =有三个不相等的实数根,求实数t 的取值范围.

一、填空题:本大题共14小题,每小题5分,共70分。

1. {|2}x x <-

2. 0

3.

4. 3

5.

6. sin(4)3

y x π

=+ 7. 10m -≤< 8. (4,2)-- 9.

10. 11. 32a ≤< 12. 8

13. 14. 9 15、解:(1)2

2cos 12sin 23)(x

x x f ωω++

=

………………………2分 .2

1

62sin +??? ?

?

+=πωx ………………………………3分

.1,22,0,=∴=∴

>=ωπω

π

ωπT …………………………………4分 令,,22

6

222

Z k k x k ∈+≤

+

≤+-

ππ

π

ππ

…………………………5分

所以,)(x f 的单调增区间为:.,6,3Z k k k ∈??

?

???++-

ππππ………………7分

(2) 2

162sin )(+??? ?

?

+

=πωx x f 的一条对称轴方程为.3π

.,2

6

3

2z k k ∈+=

+

?

∴ππ

π

π

ω…………………9分

.2

1

23+=

∴k ω…………………11分 又20<<ω,∴.131

<<-k

.2

1

,0=∴=∴ωk …………………14分

16、解 (1)因为→BC ?→CA =→CA ?→

AB ,则()()BC BA BC BA BC BA -=-,所以22BC BA =,

即||||BC BA =.

所以△ABC 是等腰三角形.

(2)因为s →//t →

,则2

2sin (2cos 1)22

C

C C -=?2sin cos 2C C C =

sin 22C C ?=sin 20C ?=,因为(0,)C π∈,则2

C π

=

因为1sin 3A =

,2C π=,则sin cos B A ==1

cos sin 3

B A ==

所以1sin(

)sin 3

2B B B π

-=

-=

17、解 (1)以O 为原点,OA 为x 轴建立直角坐标系,则(1,0),(0,1),(A B C -

设(,0)(01)D t t ≤≤,则(OC OD t +=-

所以||(OC OD t +=,当t =

min ||OC OD +=. (2)由题意1

1(,0),(0,)22D E -,设(cos ,sin )C θθ,3[0,]2

θπ∈所以

1111(cos ,sin )(cos ,sin )1sin cos )1

22224CE CD π

θθθθθθθ?=-+=+-=-+

因为3

[0,]2

θπ∈,则1sin()[42πθ-∈-,所以3[1]2CE CD ?∈-.

18、(1)由题意,当MN 和AB 之间的距离为1米时,MN 应位于DC 上方,且此时

△EMN 中MN 边上的高为0.5米,又因为EM=EN=1米,所以MN=

米,所以

,即三角通风窗EMN 的通风面积为

(2)当

MN

在矩形区域内滑动,即

时,△EMN

的面积

当MN 在半圆形区域内滑动,即

时,△EMN 的面积

综上可得;

(3)当MN 在矩形区域内滑动时,f (x )在区间上单调递减,则f (x )

<f (0)=

;当

MN

在半圆形区域内滑动,

等号成立时

因此

(米)时,每个三角形得到最大通风面积

为平方米.

19、解:设∠CPQ=θ,则CP=PQcos θ,CQ=PQsin θ (1)

∴ ∴

(2)分别以AB ,AD 所在直线为x 轴、y 轴建立平面直角坐标系, 设Q (x ,1),P (1,y ),设∠DAQ=α,∠PAB=β ∴

,即xy+(x+y )=1

又tan α=x ,tan β=y ∴,

20、解 (1)2

2(2),,

()2(2),,

x a x x a f x x x a x x a x x a ?+-?=-+=?-++

由()f x 在R 上是增函数,则2,22,2a a a a -?-???+???

≥≤即22a -≤≤,所以a 的取值范围为22a -≤≤.

(2)由题意得对任意的实数[1,2]x ∈,()()f x g x <恒成立,即1x x a -<,当[1,2]x ∈恒成立,

即1x a x -<

,11x a x x -<-<,得11

x a x x x -<<+, 故只要1x a x -<且1

a x x

<+在[1,2]x ∈上恒成立即可,

在[1,2]x ∈时,只要1x x -的最大值小于a 且1

x x

+的最小值大于a 即可,

而当[1,2]x ∈时,21110x x x '??-=+> ???,1x x -为增函数,max 132x x ?

?-= ??

?;

当[1,2]x ∈时,21110x x x '??+=-> ???,1x x +为增函数,min 12x x ?

?+= ??

?,所以322a <<.

(3)当22a -≤≤时,()f x 在R 上是增函数,则关于x 的方程()()f x t f a =不可能有三个不等的实数根;

则当(2,4]a ∈时,由2

2(2),,

()(2),x a x x a f x x a x x a

?+-?=?-++

得在x a ≥时,2()(2)f x x a x =+-对称轴2

2

a x a -=<,则()f x 在[,)x a ∈+∞为增函数,此时()f x 的值域为[(),)[2,)f a a +∞=+∞, 在x a <时,2()(2)f x x a x =-++对称轴22a x a +=

<,则()f x 在2,2a x +?

?∈-∞ ??

?为增函数,此时()f x 的值域为2(2),4a ??+-∞ ???,()f x 在2,2a x a +??∈????为减函数,此时()f x 的值域为2(2)2,4a a ??

+ ???

由存在(2,4]a ∈,方程()()2f x t f a ta ==有三个不相等的实根,则2(2)22,4a ta a ??

+∈ ???,

即存在(2,4]a ∈,使得2(2)1,8a t a ??+∈ ???

即可,令2(2)14()488a g a a a a +??

==++ ???,

只要使()max ()t g a <即可,而()g a 在(2,4]a ∈上是增函数,()max 9

()(4)8

g a g ==, 故实数t 的取值范围为91,

8??

???

; 同理可求当[4,2)a ∈--时,t 的取值范围为91,8?? ???

2017年四川省绵阳市高一上学期期末数学试卷与解析答案

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