【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�;喖纯桑�
(Ⅱ)先表示出f(x),然后化簡,對λ分類[0,1]和(1,+∞)根據(jù)最大值,確定λ的值.
解答:解:(Ⅰ)
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=cos2x(2分)
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=
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=
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(5分)
因?yàn)閤∈
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,所以cosx≥0所以|
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|=2cosx(6分)
(Ⅱ)f(x)=
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-2 λ|
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|=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=
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⇒λ=
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(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=
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⇒
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<1不合題意,舍去.(12分)
綜上,所以 λ=
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(13分)
點(diǎn)評: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.