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2019-2020年高一上学期期末统考数学试题 含答案

2019-2020年高一上学期期末统考数学试题 含答案
2019-2020年高一上学期期末统考数学试题 含答案

2019-2020年高一上学期期末统考数学试题 含答案

2017.1

(满分160分,考试时间120分钟)

注意事项:

1. 答卷前,请考生务必将自己的学校、姓名、考试号等信息填写在答卷规定的地方. 2.试题答案均写在答题卷相应位置,答在其它地方无效.

一、填空题(本大题共14小题,每小题5分,共70分,请将答案填写在答题卷相应的位置上) 1.4tan

3

π

= ▲ . 2.计算:2lg 2lg 25+= ▲ .

3.若幂函数()f x x α=的图象过点(4,2),则(9)f = ▲ . 4.已知角α的终边经过点(2,)(0)P m m >

,且cos α=

,则m = ▲ . 5.在用二分法求方程3210x x --=的一个近似解时,现在已经将一根确定在区间(1,2)内,则下一步可断定该根所在的区间为 ▲ .

6.某扇形的圆心角为2弧度,周长为4 cm ,则该扇形面积为 ▲ cm 2. 7.若3a b +=,则代数式339a b ab ++的值为 ▲ .

8.已知0.6log 5a =,45

2b =,sin1c =,将,,a b c 按从小到大的顺序用不等号“<”连接为 ▲ .

9.将正弦曲线sin y x =上所有的点向右平移2

3π个单位长度,再将图象上所有点的横坐标

变为原来的1

3倍(纵坐标不变),则所得到的图象的函数解析式y = ▲ .

10.已知函数()f x 为偶函数,且(2)()f x f x +=-,当(0,1)x ∈时,1()()2x f x =,则7

()2

f =

▲ .

11.已知21

()ax x f x x

++=在[2,)+∞上是单调增函数,则实数a 的取值范围为 ▲ .

12.如图所示,在平行四边形ABCD 中,4AB =,3AD =,E 是边CD 的中点,1

3

D F D A

=

,若4A E B F ?=-,则E

F D

C

B

A

(第12题)

sin BAD ∠= ▲ .

13.已知12(1)()32(1)

x x f x x x -?≥=?-

(cos sin )032f θλθ+-+>恒

成立,整数λ的最小值为 ▲ .

14.已知函数1

()ln()f x a x =-(a R ∈).若关于x 的方程ln[(4)25]()0a x a f x -+--=的解

集中恰好有一个元素,则实数a 的取值范围为 ▲ .

二、解答题:(本大题共6道题,计90分.解答应写出必要的文字说明、证明过程或演算步骤)

15.(本题满分14分)

已知全集U R =,集合{|27}A x x =≤<,3{|0log 2}B x x =<<,{|1}C x a x a =<<+. (1)求A

B ,()

U C A B ;

(2)如果A C =?,求实数a 的取值范围.

16.(本题满分14分)

已知:θ为第一象限角,(sin(),1)a θπ=-,1(sin(),)22

b πθ=--.

(1)若//a b ,求

sin 3cos sin cos θθ

θθ

+-的值;

(2)若||1a b +=,求sin cos θθ+的值.

17.(本题满分14分)

某工厂生产甲、乙两种产品所得利润分别为P 和Q (万元),它们与投入资金m (万

150万元资金投入生产甲、乙两

种产品,并要求对甲,乙两种产品的投资金额不低于25万元.

(1)设对乙产品投入资金x 万元,求总利润y (万元)关于x 的函数关系式及其定义域; (2)如何分配使用资金,才能使所得总利润最大?最大利润为多少?

18.(本题满分16分)

已知函数)(0)4y x π

ωω=+>.

(1)若4

π

ω=

,求函数的单调增区间和对称中心;

(2)函数的图象上有如图所示的,,A B C 三点,且满足AB BC ⊥. ①求ω的值;

②求函数在[0,2]x ∈上的最大值,并求此时x 的值.

19.(本题满分16分)

已知函数1

()1

x x e f x e -=+(e 为自然对数的底数,2,71828

e =).

(1)证明:函数()f x 为奇函数;

(2)判断并证明函数()f x 的单调性,再根据结论确定23

(1)()4f m m f -++-与0的大小关

系;

(3)是否存在实数k ,使得函数()f x 在定义域[,]a b 上的值域为[,]a b ke ke .若存在,求出实

数k 的取值范围;若不存在,请说明理由.

20.(本题满分16分)

设函数2

()||2

f x ax x b

=-+(a,b R

∈).

(1)当

15

2,

2

a b

=-=-时,解方程(2)0

x

f=;

(2)当0

b=时,若不等式()2

f x x

≤在[0,2]

x∈上恒成立,求实数a的取值范围;(3)若a为常数,且函数()

f x在区间[0,2]上存在零点,求实数b的取值范围.

扬州市2016—2017学年度第一学期期末调研测试试题

高 一 数 学 参 考 答 案 2017.1

1 2.

2 3.

3 4.1 5.3

(,2)2 6. 1 7.27

8. a c b << 9.2sin(3)3πy x =- 10 11.1

[,)4

+∞ 12.4 13.1 14.(1,2]{3,4}

15.解:(1)由30log 2x <<,得19x << ∴{|19}B x x =<<, ∴(1,9)A

B =, ............4分

(,2)[7,)U C A =-∞+∞,()

(1,2)[7,9)U C A B =; ............8分

(2)A C =? ∴12a +≤或7a ≥,解得:1a ≤或7a ≥. ............14分

16.解:(1)(sin(),1)(sin ,1)a θπθ=-=-,1

(cos ,)2

b θ=-

//a b ∴1

cos sin 02θθ-=, 化简得:tan 2θ=(不求也可以), ...........4分

sin 3cos tan 3

5sin cos tan 1

θθθθθθ++==-- ...........7分

(2)||1a b += ∴21(sin cos )14θθ-++=,则1

sin cos 8

θθ= ............11分

25

(sin cos )12sin cos 4

θθθθ∴+=+=

θ为第一象限角 sin 0,cos 0θθ∴>>,则sin cos θθ+=

............14分 17.解:(1)对乙产品投入资金x 万元,则对甲产品投入资金(150x -)万元;

所以11

(150)657619133y P Q x x =+=-+++-+, ............5分

25150150

25150x x ≤-≤??

≤≤?

,解得:25125x ≤≤,∴其定义域为[25,125]; ............7分

(2)令t 则[5,5]

t ∈,则原函数化为关于t 的函数:21

()41913

h t t t =-++,

t ∈ .............10分 所以当6t =,即36x =时,max max ()(6)203y h t h ===(万元)

答:当对甲产品投入资金114万元,对乙产品投入资金36万元时,所得总利润最大,最大利

润为203万

元. .............14分 18.解:(1)3sin()44

y x ππ

=+.

22,2

4

4

2

k x k k Z π

π

π

π

ππ-

+≤

+

+∈,解得:3818,k x k k Z -+≤≤+∈

∴函数的单调增区间为[38,18]()k k k Z -++∈; .............4分

,4

4

x k k Z π

π

π+

=∈ 14,x k k Z ∴=-+∈ ∴函数的对称中心为

(14,0)()k k Z -+∈.............8分

(2)①由图知:点B 是函数图象的最高点,设0(B x ,函数最小正周期为T ,则

003(,0),(,0)44T T A x C x -

+ 3(,3),(3

)4

4

T

T

AB BC ∴==

, ............10分 AB BC ⊥ 233016AB BC T ∴?=

-=,解得:4T = 242

ππω∴==. ............12分

[0,2]x ∈ 5[,]2

444

x π

π

ππ

+

∈ s i n (),1]

24x ππ∴+∈

∴函数在[0,2] ............14分 此时

2,2

4

2

x k k Z π

π

π

π+

=

+∈,则1

4,2

x k k Z =

+∈; [0,2]x ∈ 1

2

x ∴=

............16分 19.解:(1)函数()f x 定义域为R , .............1分

对于任意的x R ∈,都有11()()11x x x x

e e

f x f x e e -----===-++,

所以函数()f x 为奇函数. .............4分 (2)在R 上任取12,x x ,且12x x <, 1212121212112()()()11(1)(1)

x x x x x x x x e e e e f x f x e e e e ----=-=++++

12x x < 120x x e e ∴<<12120,10,10x x x x e e e e ∴-<+>+>

12()()0f x f x ∴-<,即12()()f x f x < ()f x ∴为R 上的增函数 .............7分

22133

1()244m m m -+=-+≥

23

(1)()

4

f m m f ∴-+≥

223333

(1)()(1)()()()04444f m m f f m m f f f ∴-++-=-+-≥-=. ............10分

(3)

()f x 为R 上的增函数且函数()f x 在定义域[,]a b 上的值域[,]a b ke ke

∴0k >且()()a b

f a ke f b ke

?=?=? 1

1x x x e ke e -∴=+在R 上有两个不等实根; .............12分 令,(0,)x t e t =∈+∞且单调增,问题即为方程2(1)10kt k t +-+=在(0,)+∞上有两个不等实根, 设2()(1)1h t kt k t =+-+,则2(1)4010(0)10

k k k k h ?-->?

-?

->??

=>??

,解得:03k <<-. .............16分 20.解:(1)当15

2,2

a b =-=-

时,2()|2|15f x x x =+-,所以方程即为:|2(22)|150x x +-= 解得:23x =或25x =-(舍),所以2log 3x =; .............3分 (2)当0b =时,若不等式||2x a x x -≤在[0,2]x ∈上恒成立;

当0x =时,不等式恒成立,则a R ∈; .............5分 当02x <≤时,则||2a x -≤在(0,2]上恒成立,即22x a -≤-≤在(0,2]上恒成立, 因为y x a =-在(0,2]上单调增,max 2y a =-,min y a >-,则222a a -≤??-≥-?

,解得:02a ≤≤;

则实数a 的取值范围为[0,2]; .............8分 (3)函数()f x 在[0,2]上存在零点,即方程||2x a x b -=-在[0,2]上有解; 设22()

()()x ax x a h x x ax x a ?-≥=?-+

当0a ≤时,则2(),[0,2]h x x ax x =-∈,且()h x 在[0,2]上单调增,所以min ()(0)0h x h ==,max ()(2)42h x h a ==-,则当0242b a ≤-≤-时,原方程有解,则20a b -≤≤;............10分

当0a >时,22()()()

x ax x a h x x ax x a ?-≥=?-+

a 上单调减,在[,)

a +∞上单调增; ① 当

22

a

≥,即4a ≥时,max min ()(2)24,()(0)0h x h a h x h ==-==,则当0224b a ≤-≤-时,原方程有解,则20a b -≤≤;

② 当22a a <≤,即24a ≤<时,2max min ()(),()(0)024a a h x h h x h ====,则当2

024

a b ≤-≤时,

原方程有解,则2

08

a b -≤≤;

③当02

a

<<时,

2

max min

()max{(),(2)}max{,42},()(0)0

24

a a

h x h h a h x h

==-==,

2

42

4

a

a

≥-,即则42

a

-+<时,

2

max

()

4

a

h x=,则当

2

02

4

a

b

≤-≤时,原方程有解,

2

8

a

b

-≤≤;

2

42

4

a

a

<-,即则04

a

<<-+

max

()42

h x a

=-,则当0242

b a

≤-≤-时,原方程

有解,则20

a b

-≤≤;...........14分

综上,当4

a<-+b的取值范围为[2,0]

a-;

当44

a

-+≤<时,实数b的取值范围为

2

[,0]

8

a

-;

当4

a≥时,实数b的取值范围为[2,0]

a

-......................................16 分

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