高考真题精选3《函数的性质》

历年高考数学真题精选（按考点分类）

专题七 函数的性质（学生版）

一．选择题（共21小题）

1．（2017?北京）已知函数1

()3()3

x x f x =-，则()(f x )

A ．是偶函数，且在R 上是增函数

B ．是奇函数，且在R 上是增函数

C ．是偶函数，且在R 上是减函数

D ．是奇函数，且在R 上是减函数

2．（2012?天津）下列函数中，既是偶函数，又在区间(1,2)内是增函数的为( ) A ．cos2y x =，x R ∈

B ．2log ||y x =，x R ∈且0x ≠

C ．,2

x x

e e y x R --=∈

D ．31y x =+，x R ∈

3．（2017?天津）已知奇函数()f x 在R 上是增函数．若21

(log )5

a f =-，2(log 4.1)

b f =，

0.8(2)c f =，则a ，b ，c 的大小关系为( ) A ．a b c <<

B ．b a c <<

C ．c b a <<

D ．c a b <<

4．（2015?天津）已知定义在R 上的函数||()21(x m f x m -=-为实数）为偶函数，记0.5(log 3)a f =，2(log 5)b f =，(2)c f m =，则a ，b ，c 的大小关系为( )

A ．a b c <<

B ．a c b <<

C ．c a b <<

D ．c b a <<

5．（2013?天津）已知函数()f x 是定义在R 上的偶函数，且在区间[0，)+∞上单调递增，若实数a 满足212

(log )(log )2f a f a f +…（1），则a 的取值范围是( )

A ．1[,2]2

B ．[1，2]

C ．1

(0,)2

D ．(0，2]

6．（2009?山东）已知定义在R 上的奇函数()f x ，满足(4)()f x f x -=-且在区间[0，2]上是增函数，则( ) A ．(25)(80)(11)f f f -<< B ．(80)(11)(25)f f f <<-

C ．(11)(80)(25)f f f <<-

D ．(25)(11)(80)f f f -<<

7．（2009?陕西）定义在R 上的偶函数()f x 满足：对任意的1x ，2(x ∈-∞，120]()x x ≠，有2121()(()())0x x f x f x -->．则当*n N ∈时，有( )

A ．()(1)(1)f n f n f n -<-<+

B ．(1)()(1)f n f n f n -<-<+

C ．(1)()(1)f n f n f n +<-<-

D ．(1)(1)()f n f n f n +<-<-

8．（2008?全国卷Ⅰ）设奇函数()f x 在(0,)+∞上为增函数，且f （1）0=，则不等式

()()

0f x f x x

--<的解集为( )

A ．(1-，0)(1?，)+∞

B ．(-∞，1)(0-?，1)

C ．(-∞，1)(1-?，)+∞

D ．(1-，0)(0?，1)

9．（2016?山东）已知函数()f x 的定义域为R ，当0x <时，3()1f x x =-；当11x -剟时，()()f x f x -=-；当12x >

时，11

()()22

f x f x +=-，则f （6）(= ) A ．2- B ．1 C ．0 D ．2

10．（2013?湖北）x 为实数，[]x 表示不超过x 的最大整数，则函数()[]f x x x =-在R 上为(

)

A ．奇函数

B ．偶函数

C ．增函数

D ．周期函数

11．（2009?重庆）已知函数()f x 周期为4，且当(1x ∈-，3]时，(1,1]

()1|2|,(1,3]

x f x x x ??∈-=?--∈??，

其中0m >．若方程3()f x x =恰有5个实数解，则m 的取值范围为( )

A ．，8

)3

B ．

C ．4(3，8)3

D ．4

(3

12．（2004?天津）定义在R 上的函数()f x 既是偶函数又是周期函数．若()f x 的最小正周期

是π，且当[0x ∈，]2π时，()sin f x x =，则5()3

f π

的值为( )

A ．1

2

-

B ．

12

C ．

D 13．（2018?全国）2()(32)f x ln x x =-+的递增区间是( ) A ．(,1)-∞

B ．3

(1,)2

C ．3

(2

，)+∞

D ．(2,)+∞

14．（2015?全国）设函数212

log (45)y x x =++在区间(,)a +∞是减函数，则a 的最小值为(

) A ．2

B ．1

C ．1-

D ．2-

15．（2016?新课标Ⅱ）已知函数()()f x x R ∈满足()(2)f x f x =-，若函数2|23|y x x =--与

()y f x =图象的交点为1(x ，1)y ，2(x ，2)y ，?，(m x ，)m y ，则1(m

i i x ==∑ )

A ．0

B ．m

C ．2m

D ．4m

16．（2017?山东）若函数()( 2.71828x e f x e =?是自然对数的底数）在()f x 的定义域上单调递增，则称函数()f x 具有M 性质，下列函数中具有M 性质的是( ) A ．()2x f x -=

B ．2()f x x =

C ．()3x f x -=

D ．()cos f x x =

17．（2016?天津）已知()f x 是定义在R 上的偶函数，且在区间(,0)-∞上单调递增，若实数a

满足|1|(2)(a f f ->，则a 的取值范围是( ) A ．1(,)2

-∞

B ．(-∞，13

)(22

?，)+∞

C ．1(2，3

)2

D ．3

(2

，)+∞

18．（2013?天津）已知函数()(1||)f x x a x =+．设关于x 的不等式()()f x a f x +<的解集为A ，若11

[,]22

A -?，则实数a 的取值范围是( )

A ．

B ．

C ．U

D ．(-∞ 19．（2017?新课标Ⅰ）函数()f x 在(,)-∞+∞单调递减，且为奇函数．若f （1）1=-，则满足1(2)1f x --剟的x 的取值范围是( ) A ．[2-，2]

B ．[1-，1]

C ．[0，4]

D ．[1，3]

20．（2017?全国）函数()y f x =的图象与函数(1)y ln x =-的图象关于y 轴对称，则()(f x =

)

A ．(1)ln x --

B ．(1)ln x -+

C ．(1)ln x --

D ．(1)ln x +

21．（2016?新课标Ⅱ）已知函数()()f x x R ∈满足()2()f x f x -=-，若函数1

x y x

+=

与()y f x =图象的交点为1(x ，1)y ，2(x ，2)y ，?，(m x ，)m y ，则1

()(m

i i i x y =+=∑ )

A ．0

B ．m

C ．2m

D ．4m

二．填空题（共8小题）

22．（2016?天津）已知()f x 是定义在R 上的偶函数，且在区间(,0)-∞上单调递增，若实数a

满足|1|(2)(a f f ->，则a 的取值范围是 ．

23．（2014?新课标Ⅱ）已知偶函数()f x 在[0，)+∞单调递减，f （2）0=，若(1)0f x ->，则x 的取值范围是 ．

24．（2016?全国）定义域为R 的偶函数()f x 为周期函数，其周期为8，当[4x ∈-，0]时，

()1f x x =+，则(25)f = ．

25．（2012?江苏）设()f x 是定义在R 上且周期为2的函数，在区间[1-，1]上，

1,10

()2,011

ax x f x bx x x +-

+??+?…剟其中a ，b R ∈．若13

()()22

f f =，则3a b +的值为 ．

历年高考数学真题精选（按考点分类）

专题七 函数的性质（教师版）

一．选择题（共21小题）

1．（2017?北京）已知函数1

()3()3

x x f x =-，则()(f x )

A ．是偶函数，且在R 上是增函数

B ．是奇函数，且在R 上是增函数

C ．是偶函数，且在R 上是减函数

D ．是奇函数，且在R 上是减函数

【答案】B

【解析】1

()3()333x x x x f x -=-=-，()33()x x f x f x -∴-=-=-，即函数()f x 为奇函数，

又由函数3x y =为增函数，1()3x y =为减函数，故函数1

()3()3

x x f x =-为增函数，

2．（2012?天津）下列函数中，既是偶函数，又在区间(1,2)内是增函数的为( ) A ．cos2y x =，x R ∈

B ．2log ||y x =，x R ∈且0x ≠

C ．,2

x x

e e y x R --=∈

D ．31y x =+，x R ∈

【答案】B

【解析】对于A ，令()cos2y f x x ==，则()cos(2)cos2()f x x x f x -=-==，为偶函数，

而()cos2f x x =在[0，]2π上单调递减，在[2

π

，]π上单调递增，

故()cos2f x x =在(1，]2π上单调递减，在[2

π

，2)上单调递增，故排除A ；

对于B ，令2()log ||y f x x ==，x R ∈且0x ≠，同理可证()f x 为偶函数，当(1,2)x ∈时，

22()log ||log y f x x x ===，为增函数，故B 满足题意；

对于C ，令(),2

x x

e e y

f x x R --==∈，()()f x f x -=-，为奇函数，故可排除C ；

而D ，为非奇非偶函数，可排除D ；故选：B ．

3．（2017?天津）已知奇函数()f x 在R 上是增函数．若21

(log )5

a f =-，2(log 4.1)

b f =，

0.8(2)c f =，则a ，b ，c 的大小关系为( ) A ．a b c << B ．b a c <<

C ．c b a <<

D ．c a b <<

【答案】C

【解析】奇函数()f x 在R 上是增函数，221

(log )(log 5)5a f f ∴=-=，

2(log 4.1)b f =，0.8(2)c f =，

又0.822122log 4.1log 5<<<<，0.822(2)(log 4.1)(log 5)f f f ∴<<，即c b a <<．

4．（2015?天津）已知定义在R 上的函数||()21(x m f x m -=-为实数）为偶函数，记0.5(log 3)a f =，2(log 5)b f =，(2)c f m =，则a ，b ，c 的大小关系为( )

A ．a b c <<

B ．a c b <<

C ．c a b <<

D ．c b a <<

【答案】C

【解析】()f x Q 为偶函数；()()f x f x ∴-=；||||2121x m x m ---∴-=-；

||||x m x m ∴--=-；22()()x m x m --=-；0mx ∴=；0m ∴=；||()21x f x ∴=-；

()f x ∴在[0，)+∞上单调递增，并且0.52(|log 3|)(log 3)a f f ==，2(log 5)b f =，(0)c f =； 220log 3log 5<

5．（2013?天津）已知函数()f x 是定义在R 上的偶函数，且在区间[0，)+∞上单调递增，若实数a 满足212

(log )(log )2f a f a f +?（1），则a 的取值范围是( )

A ．1[,2]2

B ．[1，2]

C ．1

(0,)2

D ．(0，2]

【答案】A

【解析】因为函数()f x 是定义在R 上的偶函数，所以1222

(log )(log )(log )f a f a f a =-=，

则212

(log )(log )2f a f a f +?（1）为：2(log )f a f ?（1），

因为函数()f x 在区间[0，)+∞上单调递增，所以2|log |1a ?，解得122

a 剟，

则a 的取值范围是1

[2

，2]，故选：A ．

6．（2009?山东）已知定义在R 上的奇函数()f x ，满足(4)()f x f x -=-且在区间[0，2]上是增函数，则( ) A ．(25)(80)(11)f f f -<< B ．(80)(11)(25)f f f <<-

C ．(11)(80)(25)f f f <<-

D ．(25)(11)(80)f f f -<<

【答案】A

【解析】(4)()f x f x -=-Q ，(8)(4)()f x f x f x ∴-=--=，即函数的周期是8， 则(11)f f =（3）(34)(1)f f f =--=--=（1），(80)(0)f f =，(25)(1)f f -=-， ()f x Q 是奇函数，且在区间[0，2]上是增函数，()f x ∴在区间[2-，2]上是增函数，

(1)(0)f f f ∴-<<（1）

，即(25)(80)(11)f f f -<<，故选：A ． 7．（2009?陕西）定义在R 上的偶函数()f x 满足：对任意的1x ，2(x ∈-∞，120]()x x ≠，有2121()(()())0x x f x f x -->．则当*n N ∈时，有( )

A ．()(1)(1)f n f n f n -<-<+

B ．(1)()(1)f n f n f n -<-<+

C ．(1)()(1)f n f n f n +<-<-

D ．(1)(1)()f n f n f n +<-<-

【答案】C

【解析】1x ，2(x ∈-∞，120]()x x ≠，有2121()(()())0x x f x f x --> 21x x ∴>时，21()()f x f x >()f x ∴在(-∞，0]为增函数

()f x Q 为偶函数()f x ∴在(0,)+∞为减函数，而110n n n +>>-…

， (1)()(1)f n f n f n ∴+<<-，(1)()(1)f n f n f n ∴+<-<-

8．（2008?全国卷Ⅰ）设奇函数()f x 在(0,)+∞上为增函数，且f （1）0=，则不等式

()()

0f x f x x

--<的解集为( )

A ．(1-，0)(1?，)+∞

B ．(-∞，1)(0-?，1)

C ．(-∞，1)(1-?，)+∞

D ．(1-，0)(0?，1)

【答案】D

【解析】由奇函数()f x 可知

()()2()

0f x f x f x x x

--=<，即x 与()f x 异号，

而f （1）0=，则(1)f f -=-（1）0=，

又()f x 在(0,)+∞上为增函数，则奇函数()f x 在(,0)-∞上也为增函数， 当01x <<时，()f x f <（1）0=，得()

0f x x

<，满足； 当1x >时，()f x f >（1）0=，得

()

0f x x

>，不满足，舍去； 当10x -<<时，()(1)0f x f >-=，得()

0f x x

<，满足； 当1x <-时，()(1)0f x f <-=，得

()

0f x x

>，不满足，舍去； 所以x 的取值范围是10x -<<或01x <<．故选：D ．

9．（2016?山东）已知函数()f x 的定义域为R ，当0x <时，3()1f x x =-；当11x -剟时，()()f x f x -=-；当12x >

时，11

()()22

f x f x +=-，则f （6）(= ) A ．2- B ．1 C ．0 D ．2

【答案】D 【解析】Q 当12x >

时，11()()22

f x f x +=-， ∴当1

2

x >

时，(1)()f x f x +=，即周期为1．f ∴（6）f =（1）， Q 当11x -剟时，()()f x f x -=-，f ∴（1）(1)f =--，

Q 当0x <时，3()1f x x =-，(1)2f ∴-=-，f ∴（1）(1)2f =--=，f ∴（6）2=．

10．（2013?湖北）x 为实数，[]x 表示不超过x 的最大整数，则函数()[]f x x x =-在R 上为(

)

A ．奇函数

B ．偶函数

C ．增函数

D ．周期函数

【答案】D

【解析】()[]f x x x =-Q ，(1)(1)[1]1[]1[]()f x x x x x x x f x ∴+=+-+=+--=-=， ()[]f x x x ∴=-在R 上为周期是1的函数．故选：D ．

11．（2009?重庆）已知函数()f x 周期为4，且当(1x ∈-，3]时，(1,1]()1|2|,(1,3]x f x x x ??∈-=?--∈??

，

其中0m >．若方程3()f x x =恰有5个实数解，则m 的取值范围为( )

A ．，8

)3

B ．

C ．4(3，8)3

D ．4

(3

【答案】B

【解析】Q 当(1x ∈-，1]时，将函数化为方程2

2

21(0)y x y m

+=…，

∴实质上为一个半椭圆，其图象如图所示，

同时在坐标系中作出当(1x ∈，3]得图象，再根据周期性作出函数其它部分的图象，

由图易知直线3

x y =与第二个椭圆22

2(4)11(0)y x y m -+==…相交，

而与第三个半椭圆2

2

2(8)11y x m -+==(0)y …无公共点时，方程恰有5个实数解，

将3

x y =代入22

2(4)11y x m -+==(0)y …得，

2222(91)721350m x m x m +-+=，令29(0)t m t =>， 则2(1)8150t x tx t +-+=，由△2(8)415t t =-?(1)0t +>，得15t >，由2915m >，且0m >得

m ，

同样由3

x y =与第三个椭圆22

2(8)11y x m -+==(0)y …由△0<可计算得m <，

综上可知m ∈

故选：B ．

12．（2004?天津）定义在R 上的函数()f x 既是偶函数又是周期函数．若()f x 的最小正周期

是π，且当[0x ∈，]2π时，()sin f x x =，则5()3

f π

的值为( )

A ．1

2

-

B ．

12

C ．

D 【答案】D

【解析】()f x Q 的最小正周期是π55()(2)()333

f f f ππππ∴=-=-

Q 函数()f x 是偶函数5(

)()sin 333f f πππ∴===

．故选：D ． 13．（2018?全国）2()(32)f x ln x x =-+的递增区间是( ) A ．(,1)-∞ B ．3(1,)2

C ．3

(2

，)+∞

D ．(2,)+∞

【答案】D

【解析】令232(1)(2)0t x x x x =-+=-->，求得1x <或2x >，

故函数的定义域为{|1x x <或2x >}，()f x lnt =，本题即求函数t 在定义域内的增区间． 结合二次函数的性质可得函数t 在定义域内的增区间为(2,)+∞，故选：D ．

14．（2015?全国）设函数212

log (45)y x x =++在区间(,)a +∞是减函数，则a 的最小值为(

) A ．2 B ．1

C ．1-

D ．2-

【答案】D

【解析】可令245t x x =++，由12

()log f x t =在(0,)+∞递减，

可得245t x x =++在(,)a +∞是增函数，且0t >在(,)a +∞恒成立，

可得2a -?且2450a a ++…，解得2a -…，则a 的最小值是2-．故选：D ．

15．（2016?新课标Ⅱ）已知函数()()f x x R ∈满足()(2)f x f x =-，若函数2|23|y x x =--与

()y f x =图象的交点为1(x ，1)y ，2(x ，2)y ，?，(m x ，)m y ，则1(m

i i x ==∑ )

A ．0

B ．m

C ．2m

D ．4m

【答案】B

【解析】Q 函数()()f x x R ∈满足()(2)f x f x =-，

故函数()f x 的图象关于直线1x =对称，函数2|23|y x x =--的图象也关于直线1x =对称， 故函数2

|23|y x x =--与()y f x = 图象的交点也关于直线1x =对称，故122

m

i i m

x m ==

?=∑，

故选：B ．

16．（2017?山东）若函数()( 2.71828x e f x e =?是自然对数的底数）在()f x 的定义域上单调递增，则称函数()f x 具有M 性质，下列函数中具有M 性质的是( ) A ．()2x f x -= B ．2()f x x = C ．()3x f x -=

D ．()cos f x x =

【答案】A

【解析】当()2x f x -=时，函数()()2

x x e

e f x =在R 上单调递增，函数()f x 具有M 性质，

17．（2016?天津）已知()f x 是定义在R 上的偶函数，且在区间(,0)-∞上单调递增，若实数a

满足|1|(2)(a f f ->，则a 的取值范围是( ) A ．1(,)2

-∞

B ．(-∞，13

)(22

?，)+∞

C ．1(2，3

)2

D ．3

(2

，)+∞

【答案】C

【解析】()f x Q 是定义在R 上的偶函数，且在区间(,0)-∞上单调递增， ()f x ∴在(0,)+∞上单调递减． |1|

2

0a ->Q

，(f f =

，1|1|

2

2

2a -∴<．1|1|2a ∴-<

，解得13

22

a <<．故选：C ． 18．（2013?天津）已知函数()(1||)f x x a x =+．设关于x 的不等式()()f x a f x +<的解集为A ，

若11

[,]22

A -?，则实数a 的取值范围是( )

A ．

B ．

C ．U

D ．(-∞ 【答案】A

【解析】取12a =-时，1()||2f x x x x =-+，()()f x a f x +

()||1||22x x x x ∴--+>，

（1）0x <时，解得304x -<<；（2）102x 剟时，解得102x

剟；（3）12x >时，解得15

24x <<， 综上知，12a =-时，3(4A =-，5

)4

，符合题意，排除B 、D ；取1a =时，()||f x x x x =+，

()()f x a f x +

（1）1x <-时，解得0x >，矛盾；（2）10x -剟，解得0x <，矛盾；

（3）0x >时，解得1x <-，矛盾；综上，1a =，A =?，不合题意，排除C ，故选：A ． 19．（2017?新课标Ⅰ）函数()f x 在(,)-∞+∞单调递减，且为奇函数．若f （1）1=-，则满足1(2)1f x --剟的x 的取值范围是( ) A ．[2-，2] B ．[1-，1]

C ．[0，4]

D ．[1，3]

【答案】D

【解析】Q 函数()f x 为奇函数．

若f （1）1=-，则(1)1f -=，又Q 函数()f x 在(,)-∞+∞单调递减，1(2)1f x --剟，

f ∴（1）(2)(1)f x f --剟，121x ∴--剟

，解得：[1x ∈，3]， 故选：D ．

20．（2017?全国）函数()y f x =的图象与函数(1)y ln x =-的图象关于y 轴对称，则()(f x =

)

A ．(1)ln x --

B ．(1)ln x -+

C ．(1)ln x --

D ．(1)ln x +

【答案】C

【解析】根据题意，函数()y f x =的图象与函数(1)y ln x =-的图象关于y 轴对称， 则有()(1)f x ln x -=-，则()(1)f x ln x =--；故选：C ．

21．（2016?新课标Ⅱ）已知函数()()f x x R ∈满足()2()f x f x -=-，若函数1

x y x

+=

与()y f x =图象的交点为1(x ，1)y ，2(x ，2)y ，?，(m x ，)m y ，则1

()(m

i i i x y =+=∑ )

A ．0

B ．m

C ．2m

D ．4m

【答案】B

【解析】函数()()f x x R ∈满足()2()f x f x -=-，即为()()2f x f x +-=， 可得()f x 关于点(0,1)对称，函数1x y x +=

，即1

1y x

=+的图象关于点(0,1)对称， 即有1(x ，1)y 为交点，即有1(x -，12)y -也为交点， 2(x ，2)y 为交点，即有2(x -，22)y -也为交点，

?

则有11221()()()()m

i i m m i x y x y x y x y =+=++++?++∑

111122221

[()(2)()(2)()(2)]2

m m m m x y x y x y x y x y x y =++-+-+++-+-+?+++-+-m =． 二．填空题（共8小题）

22．（2016?天津）已知()f x 是定义在R 上的偶函数，且在区间(,0)-∞上单调递增，若实数a

满足|1|(2)(a f f ->，则a 的取值范围是 ． 【答案】1(2，3)2

【解析】()f x Q 是定义在R 上的偶函数，且在区间(,0)-∞上单调递增，

()f x ∴在区间[0，)+∞上单调递减，则|1|(2)(a f f ->，等价为|1|(2)a f f ->，

即|1|2a -<1|1|2a -<

，即1322a <<，故答案为：1(2，3

)2

23．（2014?新课标Ⅱ）已知偶函数()f x 在[0，)+∞单调递减，f （2）0=，若(1)0f x ->，则x 的取值范围是 ． 【答案】(1,3)-

【解析】Q 偶函数()f x 在[0，)+∞单调递减，f （2）0=，

∴不等式(1)0f x ->等价为(1)f x f ->（2），即(|1|)f x f ->（2），|1|2x ∴-<，

解得13x -<<，故答案为：(1,3)-

24．（2016?全国）定义域为R 的偶函数()f x 为周期函数，其周期为8，当[4x ∈-，0]时，

()1f x x =+，则(25)f = ．

【答案】0

【解析】Q 定义域为R 的偶函数()f x 为周期函数，其周期为8，

当[4x ∈-，0]时，()1f x x =+，(25)(831)f f f ∴=?+=（1）(1)110f =-=-+=． 故答案为：0．

25．（2012?江苏）设()f x 是定义在R 上且周期为2的函数，在区间[1-，1]上，

1,10

()2,011

ax x f x bx x x +-

+??+??剟其中a ，b R ∈．若13

()()22

f f =，则3a b +的值为 ．

【答案】10-

【解析】()f x Q 是定义在R 上且周期为2的函数，1,10

()2,011

ax x f x bx x x +-

=?

+??+??剟，

311()()1222f f a ∴=-=-，14()23b f +=

；又13()()22f f =，14

123

b a +∴-=① 又(1)f f -=（1），20a b ∴+=，②

由①②解得2a =，4b =-；310a b ∴+=-．故答案为：10-．

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