函数f(x)=3cos2x-sin2x的单调减区间为()A.[kπ+π6,π+2π3],k∈ZB.[kπ-7π12,π-π12],k∈ZC.[2kπ-7π12,2kπ-π12],k∈ZD.[kπ-π1

题目简介

函数f(x)=3cos2x-sin2x的单调减区间为()A.[kπ+π6,π+2π3],k∈ZB.[kπ-7π12,π-π12],k∈ZC.[2kπ-7π12,2kπ-π12],k∈ZD.[kπ-π1

题目详情

函数f(x)=
3
cos2x-sin2x的单调减区间为(  )
A.[kπ+
π
6
,π+
3
],k∈Z
B.[kπ-
12
,π-
π
12
],k∈Z
C.[2kπ-
12
,2kπ-
π
12
],k∈Z
D.[kπ-
π
12
,kπ+
12
],k∈Z
题型:单选题难度:偏易来源:不详

答案

∵函数f(x)=
3
cos2x-sin2x=2(
3
2
cos2x-class="stub"1
2
sin2x)=2sin(class="stub"π
3
-2x)=-2sin(2x-class="stub"π
3
),
故本题即求y=2sin(2x-class="stub"π
3
)的增区间.
由 2kπ-class="stub"π
2
≤2x-class="stub"π
3
≤2kπ+class="stub"π
2
,k∈z,可得 kπ-class="stub"π
12
≤x≤2kπ≤kπ+class="stub"5π
12
,k∈z.
故y=2sin(2x-class="stub"π
3
)的增区间为[kπ-class="stub"π
12
,kπ+class="stub"5π
12
],k∈Z,
故选D.

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