已知函数f(x)=3sin(2x-π6)+2sin2(x-π12)(x∈R).(1)求函数f(x)的最小正周期;(2)求使函数f(x)取得最大值的x的集合;(3)求函数f(x)的单调递增区间.-数学

题目简介

已知函数f(x)=3sin(2x-π6)+2sin2(x-π12)(x∈R).(1)求函数f(x)的最小正周期;(2)求使函数f(x)取得最大值的x的集合;(3)求函数f(x)的单调递增区间.-数学

题目详情

已知函数f(x)=
3
sin(2x-
π
6
)+2sin2(x-
π
12
)
(x∈R).
(1)求函数f(x)的最小正周期;
(2)求使函数f(x)取得最大值的x的集合;
(3)求函数f(x)的单调递增区间.
题型:解答题难度:中档来源:不详

答案

(1)因为f(x)=
3
sin(2x-class="stub"π
6
)+1-cos2(x-class="stub"π
12
)

=2[
3
2
sin(2x-class="stub"π
6
)-class="stub"1
2
cos(2x-class="stub"π
6
)]+1
=2sin[(2x-class="stub"π
6
)-class="stub"π
6
]+1
=2sin(2x-class="stub"π
3
)+1

所以f(x)的最小正周期T=class="stub"2π
2

(2)当f(x)取最大值时,sin(2x-class="stub"π
3
)=1
,此时2x-class="stub"π
3
=2kπ+class="stub"π
2
(k∈Z),即x=kπ+class="stub"5π
12
(k∈Z),
所以所求x的集合为{x|x=kπ+class="stub"5π
12
}
(k∈Z).
(3)由-class="stub"π
2
+2kπ≤2x-class="stub"π
3
≤class="stub"π
2
+2kπ得,kπ-class="stub"π
12
≤x≤kπ+class="stub"5π
12
,k∈Z

函数f(x)的单调递增区间为[kπ-class="stub"π
12
,kπ+class="stub"5π
12
],k∈Z

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