【答案】
分析:(1)由奇函數(shù)的特性f(0)=0,解出a=1可得f(x)的解析式為f(x)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103173820402108927/SYS201311031738204021089020_DA/0.png)
.再由指數(shù)函數(shù)的值域,解關(guān)于y的不等式即可求出f(x)的值域;
(2)將原不等式化簡(jiǎn),可得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103173820402108927/SYS201311031738204021089020_DA/1.png)
≤
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103173820402108927/SYS201311031738204021089020_DA/2.png)
對(duì)x∈
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103173820402108927/SYS201311031738204021089020_DA/3.png)
恒成立,由此結(jié)合對(duì)數(shù)函數(shù)的單調(diào)性和定義域,化簡(jiǎn)得到k
2≤1-x
2對(duì)于x∈
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103173820402108927/SYS201311031738204021089020_DA/4.png)
恒成立,可得實(shí)數(shù)k的取值范圍.
解答:解:(1)令t=2x,得f (x)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103173820402108927/SYS201311031738204021089020_DA/5.png)
-------------------------------(1分)
∵f (x)是奇函數(shù),∴f(0)=0,解之可得a=1
∴函數(shù)的解析式為f(x)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103173820402108927/SYS201311031738204021089020_DA/6.png)
-----------------------------(3分)
∵由y=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103173820402108927/SYS201311031738204021089020_DA/7.png)
解出2
x=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103173820402108927/SYS201311031738204021089020_DA/8.png)
>0,解之得-1<y<1
∴值域?yàn)?nbsp;(-1,1)-------------------------------------------------(6分)
(2)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103173820402108927/SYS201311031738204021089020_DA/9.png)
≤
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103173820402108927/SYS201311031738204021089020_DA/10.png)
對(duì)x∈
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103173820402108927/SYS201311031738204021089020_DA/11.png)
恒成立
即:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103173820402108927/SYS201311031738204021089020_DA/12.png)
≤
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103173820402108927/SYS201311031738204021089020_DA/13.png)
,
不等式
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103173820402108927/SYS201311031738204021089020_DA/14.png)
≤
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103173820402108927/SYS201311031738204021089020_DA/15.png)
對(duì)x∈
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103173820402108927/SYS201311031738204021089020_DA/16.png)
恒成立------(8分)
即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103173820402108927/SYS201311031738204021089020_DA/17.png)
----①,對(duì)于x∈
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103173820402108927/SYS201311031738204021089020_DA/18.png)
恒成立
由①,得k
2≤1-x
2對(duì)于x∈
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103173820402108927/SYS201311031738204021089020_DA/19.png)
恒成立---------------------------(10分)
∴k
2≤1-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103173820402108927/SYS201311031738204021089020_DA/20.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103173820402108927/SYS201311031738204021089020_DA/21.png)
,解之得0<k≤
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103173820402108927/SYS201311031738204021089020_DA/22.png)
----------------------------------(12分)
點(diǎn)評(píng):本題給出含有指數(shù)式的分式型函數(shù),求函數(shù)的奇偶性和值域,并依此討論不等式恒成立時(shí)實(shí)數(shù)k的范圍.著重考查了基本初等函數(shù)的單調(diào)性、奇偶性和函數(shù)恒成立問(wèn)題等知識(shí),屬于中檔題.