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2017—2018学年第一学期高一期中考试数学试题及参考答案

2017—2018学年第一学期高一期中考试数学试题及参考答案
2017—2018学年第一学期高一期中考试数学试题及参考答案

2017--2018学年第一学期高一期中考试

数学学科试题

试卷分值:160分 考试时间:120分钟

一、填空题:本大题共14小题,每小题5分,共计70分.不需要写出解答过程,请将答案 填写在答题卡相应的位置上..

1 ?若集合 A={1, 3} , B={0, 3},则 A U B= _____ . 2.计算:sin210的值为

3?若扇形的半径为 2,圆心角为 一,则它的面积为 ____________.

3 4、 函数f x a x 1

1 a 0, a 1过定点 ________________ .

1

5、 若一个幕函数f (x)的图象过点(2.—),则f(x)的解析式为

4

6、 已知a=20.3, b=20.4, c=log 20.3,则a , b , c 按由大到小排列的结果是 _______ .

7、函数f x . 1 log 3 x 1的定义域是 4

的终边上,且满足 x <0 , cos =,则tan

5

0的解集为 . 1 -(0 ),贝U sin cos .

5

ax 2

4 a 3 x 5在区间 .2上是减函数,贝U a 的取值范围 f(

■宜一f(x2) 0成立,则实数 m 的取值范围是 _______________

13、 已知函数f x 是定义在R 上的偶函数,若 f x 在 .0上是减函数,且f 2 0,

f x

则丄冬 0的x 的取值范围为 __________________________ .

x

x.(x m)

14、 已知函数f(x) 2

,其中m>0,若存在实数b.使得关于x 的方

x 2 2mx 4m.(x m)

程f (x) b 有三个不同的根.则m 的取值范围是 ____________________ .

二、解答题:本大题共 6小题,共计90分?请在答题纸指定区域.内作答,解答应写出文字 说明,

证明过程或演算步骤

.

15. (本小题满分14分)已知集合 A={x|x 2- 2x - 8< 0},集合B m 3. m m R (1 )若A n B=[2 , 4],求实数 m 的值;

&已知点M (4. x)在角 9、不等式4x

2x 2

3

10、 已知 sin cos 11、 关于X 的函数f x

12、已知定义在R 上的函数f x

2x 1 mx m

x 0 1.x 0

满足对任意x 1

X 2都有

(2 )设全集为R,若A? ?R B,求实数m的取值范围.

16. (本小题满分14分)

2_ ________

(1

(2)( Ig5) 2+|g2?|g50.

17. (本小题满分14分)

已知y=f ( x)( x € R)是偶函数,当x > 0 时,f (x) =x2- 2x.

(1 )求f (x)的解析式;

(2)若不等式f ( x)> mx在K x< 2时都成立,求m的取值范围.

18. (本小题满分16分)已知函数f (x)= 为奇函数.

(1 )求a的值;

(2)证明:f (x)是R上的增函数;

3

(3)解不等式:flog2x <

19. (本小题满分16分)如图,在长为10千米的河流0C的一侧有一条观光带,观光带的前一部分为曲线段OAB ,设曲线段OAB为函数y ax2 bx c a 0 , x €[0, 6](单位:千米)的图象,且图象的最高点为 A (4, 4);观光带的后一部分为线段BC .

20. (本小题满分16分)若函数f x和g x满足:①在区间[a, b]上均有定义;②函数y f x g x在区间[a, b]上至少有一个零点,则称f x和g x在区间[a, b]上具有关系G.

(1 )若f x lg x, g x 3 x,试判断f x和g x在[1, 4]上是否具有关系G,并说明理由;

(2)若fx 2 x 2 1和g x mx2在[1, 4]上具有关系G,求实数m的取值范围.

2017--2018学年第一学期高一期中考试

数学学科试题(答案)

一、填空题

1、{0, 1, 3};

2、-

1

2

3

4

、3 ;

4 、1,2 ;5、f2

x x

6、b, a, c.; 7 、1,4;8

3

、—9 、0, log2 3 ;

10

7 . 45

11、[0, 3];

2

12、0m 3 ;13、,2 0,2 ;14、3,

二、解答题

15.【解答】解:(I A={x|(x+2)( :x- 4)< 0}==[ - 2,4]- 3分??? A n B=[2, 4],

解得m=5

(II)由(I)知C R B={X| x v m - 3,或x>m},

T A? C R B,A 4v m - 3,或-2>m,解得m v - 2,或m>7.

故实数m的取值范围为(-x,- 2)U( 7, +x) --------------------------------- 14分

16. 【解答】解:(1)原式=“斗-(施-D+3+1 ------------------------------- 3分

2

=4-近+1 +3+1

=9-施----------------- 7分

(2)原式=lg25+lg2 (1+lg5)

=lg5 (lg5+lg2) +lg2 -------------------- 10 分

=lg5+lg2

=1. ---------------------- 14 分

17、【解答】解:(1)当x v 0时,有-x> 0,

??? f (x)为偶函数,??? f (x) =f (- x) = (- x) 2- 2 (- x) =x2+2x, ------------- 4 分

? f (x) = . ----------------------------------------- 6分

工+2心

10分

(2)由题意得x 2-2x >mx 在 Kx <2时都成立,

即 x - 2> m 在 K x < 2 时都成立, ------------------------ 10分 即m W x - 2在 Kx < 2时都成立.

而在 1W x <2 时,(x -2) min =- 1 ,二 m W- 1. ------------------------------- 14分 18. --------------------------------------------------------------------------------- 【解答】(1)解:f (x )的定义域为 R ------------------------------------------------------------- 2分 ??? f (x )为奇函数,

二 f (-x ) = - f(x),.°. a=1. ------------------------------ 5分

设 X 1 € R, X 2€ R,且 X 1V x 2, ??? f (X 1)- f ( x 2)=」-

-=「一,. ___________________ 8 分

2 ;+1

2 +1 〔『耳 1)〔2?赳)

???尹乜勺,

二 f (X 1)— f ( x 2)V 0. 二 f (X 1)V f ( X 2).

f (Io

g 2x )w 岸即 f (log 2x )w f (2)

5

??? f (x )为R 上的增函数,

? gx W 2. -------------------------------------------------------- 15分 ? 0 V x W 4. ----------------------------------------------------- 16 分

19. 【解答】解:(1)因为曲线段OAB 过点0,且最高点为A (4, 4),

因为后一部分为线段BC, B (6, 3), C (10, 0),

(2)证明:易得f (x ) =1 -

2 2x 1

??? f (x )为R 上的增函数.

------------------------------------------------ 11分 (3)令 f (x ) 十,解得x=2. -------------------------------------

13分 所以

亡二0

L6a+4b+c=4 b 4

----- =4

2a

解得

所以,当x € [0, 6]时,

1

b-2

lx 2 2x

4

(3 分)

20. ------------------------------------------------------------------------ 【解答】解:(1)它们具有关系G--------------------------------------------- 2分 令 h (x ) =f (x )— g (x ) =lgx+x - 3, ??? h (1) =— 2v 0,h (4) =lg4+1>0;

故 h (1) ?h (4)v 0,又 h (x )在[1, 4]上连续,

故函数y=f (x )— g (x )在区间[a ,b ]上至少有一个零点, 故f (x )和g (x )在[1, 4]上具有关系G. -------------------------- 6分 (2)令 h (x ) =f (x )— g (x ) =2|x — 2|+1 — mx 2,

当m < 0时,易知h (x )在[1, 4]上不存在零点, ------------- . ,

工-宜2<工杀4

当 m >0 时,h (x ) = p 严一;

-2K +5, 2

当1< x < 2时,

由二次函数知h (x )在[1, 2]上单调递减, 故;

3 15

f x — X — 当 x € [6, 10]时,

4

2

(6 分) x 6 [0, 6] 冷岸詐xE (6, 10] (2)设 OM=t (Ovt <2),贝U

S=-^t 2+2t- PN=-^t 2+2t 综上,fg)二* (8 分)

由i_ - -zi — 所以点-, i —+ 厶丨-',「 d J

所以,绿化带的总长度 y =

MQ+Qp+pN

=_': y ' I _ - ■丄* 丄-—- 旦 丫冋- 6 当t=1时, ,得 (11 分)

(13

分)

所以,当OM 长为1千米时,绿化带的总长度最长

(16 分)

故m€ [当,3]; ------------ 11 分

当m€( 0,書)U( 3, +x)时,

若m€( 0,寺),则h (乂)在(2, 4]上单调递增,

而h (2)>0, h (4)>0;

故没有零点;------------- 13分

若m€( 3, +x),贝U h (乂)在(2, 4]上单调递减,

此时,h (2) =-4m+1v0;

故没有零点;------------- 15分

综上所述,

若 f (x) =2| x- 2|+ 1 和g (x) =mx?在[1, 4]上具有关系G, 则m€ [丄,3] . ------------------ 16 分

浙江湖州市2016-2017学年高一英语下学期期中试题

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