【答案】
分析:(Ⅰ)利用坐標(biāo)運(yùn)算求數(shù)量積,再用兩角差的余弦直求解;先求向量和,再求和的模化簡即可.
(Ⅱ)先表示出f(x),然后化簡,對(duì)λ分類[0,1]和(1,+∞)根據(jù)最大值,確定λ的值.
解答:解:(Ⅰ)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180529380390669/SYS201310241805293803906012_DA/0.png)
=cos2x(2分)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180529380390669/SYS201310241805293803906012_DA/1.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180529380390669/SYS201310241805293803906012_DA/2.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180529380390669/SYS201310241805293803906012_DA/3.png)
(5分)
因?yàn)閤∈
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180529380390669/SYS201310241805293803906012_DA/4.png)
,所以cosx≥0所以|
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180529380390669/SYS201310241805293803906012_DA/5.png)
|=2cosx(6分)
(Ⅱ)f(x)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180529380390669/SYS201310241805293803906012_DA/6.png)
-2 λ|
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180529380390669/SYS201310241805293803906012_DA/7.png)
|=cos
2x-4 λcosx=2cos
2x-4 λcosx-1
=2(cosx-λ)
2-1-2 λ
2(8分)
令t=cosx∈[0,1],則f(x)=g(t)=2(t-λ)
2-1-2λ
2①當(dāng)0≤λ≤1時(shí),當(dāng)且僅當(dāng)t=λ時(shí),f(x)取得最小值,
g( λ)=-1-2 λ
2即-1-2 λ
2=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180529380390669/SYS201310241805293803906012_DA/8.png)
⇒λ=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180529380390669/SYS201310241805293803906012_DA/9.png)
(10分)
②當(dāng) λ>1時(shí),當(dāng)且僅當(dāng)t=1時(shí),f(x)取得最小值,g(1)=1-4λ
即1-4λ=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180529380390669/SYS201310241805293803906012_DA/10.png)
⇒
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180529380390669/SYS201310241805293803906012_DA/11.png)
<1不合題意,舍去.(12分)
綜上,所以 λ=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180529380390669/SYS201310241805293803906012_DA/12.png)
(13分)
點(diǎn)評(píng):本題考查平面向量數(shù)量積的運(yùn)算,向量的模,函數(shù)最值,是中檔題.