题目内容
已知向量
(I)若


(II)设f(θ)=

【答案】分析:(I)由题设条件,
∥
,,可得sinθ=cosθ,由此得tanθ=1,再由
,即可判断出θ的值;
(II)由f(θ)=
及两向量的坐标得到f(θ)的函数解析式,再由三角函数的最值的判断出函数的最值,利用正弦函数的单调性求出函数的单调递增区间.
解答:解:(I)因为
∥
,,可得sinθ=cosθ,由此得tanθ=1,又
,故有θ=
(II)f(θ)=
=2sinθcosθ+2cos2θ+1=sin2θ+cos2θ+2=
sin(2θ+
)+2
因为θ∈
,所以2θ+
∈
∴函数f(θ)的最大值为
+2,
令
解得θ∈
故函数的单调递增区间是
点评:本题考查平面向量数量积的运算及三角函数的最值求法,解题的关键是熟练掌握向量的数量积的运算,平面向量数量积是考试的一个热点,应注意总结其运算规律,三角函数的最值在近年的高考中出现的频率也很高,在某些求最值的问题中,将问题转化到三角函数中利用三角函数的有界性求函数最值,方便了求最值



(II)由f(θ)=

解答:解:(I)因为




(II)f(θ)=



因为θ∈



∴函数f(θ)的最大值为

令

解得θ∈

故函数的单调递增区间是

点评:本题考查平面向量数量积的运算及三角函数的最值求法,解题的关键是熟练掌握向量的数量积的运算,平面向量数量积是考试的一个热点,应注意总结其运算规律,三角函数的最值在近年的高考中出现的频率也很高,在某些求最值的问题中,将问题转化到三角函数中利用三角函数的有界性求函数最值,方便了求最值

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