Question

已知函数f(x)=2cos∧2x+2√3sinxcosx求函数f（x）的最小正周期及最值和单调增期间

急用，谁帮帮我

Answer 1

解：f(x)=2cos�0�5x+2√3sinxcosx
=(2cos�0�5x-1)+1+√3×2sinxcosx
=cos2x+√3sin2x+1
=2sin(2x+π/6)+1
函数f(x)最小正周期T=2π/2=π
由2kπ≤2x+π/6≤2kπ+π/2 (k∈z)，得
kπ-π/12≤x≤kπ+π/6
或 由2kπ+3π/2≤2x+π/6≤2kπ+2π(k∈z)，得
kπ+2π/3≤x≤kπ+11π/12
因此，f(x)的单独递增区间为[kπ-π/12,kπ+π/6]或[kπ+2π/3，kπ+11π/12] (k∈z)

Answer 2

倍角公式变形化简即可求得最小正周期为派.最大值为三,最小值负一.结合三角函数的相应区间可求单调区间.

Answer 3

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