【答案】
分析:化簡函數(shù)為同角同名函數(shù),利用2cos
2x-1=cos2x,sin2x+cos2x=)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102429881839451/SYS201311031024298818394015_DA/0.png)
sin(2x+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102429881839451/SYS201311031024298818394015_DA/1.png)
).再利用正弦函數(shù)的性質(zhì),對稱軸方程x=kπ+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102429881839451/SYS201311031024298818394015_DA/2.png)
,k∈z;遞減區(qū)間為[2kπ+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102429881839451/SYS201311031024298818394015_DA/3.png)
,2kπ+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102429881839451/SYS201311031024298818394015_DA/4.png)
],k∈z,及函數(shù)圖象的變化規(guī)律解決.
解答:解:首先對函數(shù)進(jìn)行化簡,f(x)=2cos
2x+sin2x-1=sin2x+cos2x=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102429881839451/SYS201311031024298818394015_DA/5.png)
(sin2xcos
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102429881839451/SYS201311031024298818394015_DA/6.png)
+cos2xsin
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102429881839451/SYS201311031024298818394015_DA/7.png)
)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102429881839451/SYS201311031024298818394015_DA/8.png)
sin(2x+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102429881839451/SYS201311031024298818394015_DA/9.png)
).
對①,令2x+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102429881839451/SYS201311031024298818394015_DA/10.png)
=kπ+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102429881839451/SYS201311031024298818394015_DA/11.png)
,得對稱軸方程x=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102429881839451/SYS201311031024298818394015_DA/12.png)
+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102429881839451/SYS201311031024298818394015_DA/13.png)
,k∈z,∴②正確;
對①,令2kπ+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102429881839451/SYS201311031024298818394015_DA/14.png)
<2x+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102429881839451/SYS201311031024298818394015_DA/15.png)
<2kπ+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102429881839451/SYS201311031024298818394015_DA/16.png)
,得 kπ+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102429881839451/SYS201311031024298818394015_DA/17.png)
<x<kπ+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102429881839451/SYS201311031024298818394015_DA/18.png)
,k∈z.函數(shù)的遞減區(qū)間為[kπ+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102429881839451/SYS201311031024298818394015_DA/19.png)
,kπ+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102429881839451/SYS201311031024298818394015_DA/20.png)
],k∈z,∴①√;
對③,平移的單位應(yīng)是
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102429881839451/SYS201311031024298818394015_DA/21.png)
,∴③×.
對④,當(dāng)x∈[0,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102429881839451/SYS201311031024298818394015_DA/22.png)
]時f(x)單調(diào)遞增,當(dāng)x∈[
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102429881839451/SYS201311031024298818394015_DA/23.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102429881839451/SYS201311031024298818394015_DA/24.png)
]時單調(diào)遞減,f(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102429881839451/SYS201311031024298818394015_DA/25.png)
)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102429881839451/SYS201311031024298818394015_DA/26.png)
,f(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102429881839451/SYS201311031024298818394015_DA/27.png)
)=-1∴值域是[-1,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102429881839451/SYS201311031024298818394015_DA/28.png)
],∴④√.
故答案是①②④
點(diǎn)評:牢記三角函數(shù)的性質(zhì)及圖象變化規(guī)律,利用整體代入求解復(fù)合函數(shù)的對稱軸、單調(diào)區(qū)間、值域是本題的關(guān)鍵.