已知函数f(x)=2sin2x+23sinxcosx+1.求:(Ⅰ)f(x)的最小正周期;(Ⅱ)f(x)的单调递增区间;(Ⅲ)f(x)在[0,π2]上的最值.-数学

题目简介

已知函数f(x)=2sin2x+23sinxcosx+1.求:(Ⅰ)f(x)的最小正周期;(Ⅱ)f(x)的单调递增区间;(Ⅲ)f(x)在[0,π2]上的最值.-数学

题目详情

已知函数f(x)=2sin2x+2
3
sinxcosx+1.
求:
(Ⅰ)f(x)的最小正周期;
(Ⅱ)f(x)的单调递增区间;
(Ⅲ)f(x)在[0,
π
2
]
上的最值.
题型:解答题难度:中档来源:不详

答案

(Ⅰ)因为f(x)=2sin2x+2
3
sinxcosx+1
=1-cos2x+2
3
sinxcosx+1

=
3
sin2x-cos2x+2

=2sin(2x-class="stub"π
6
)+2

所以f(x)的最小正周期T=class="stub"2π
2
=π.

(Ⅱ)因为f(x)=2sin(2x-class="stub"π
6
)+2

所以由2kπ-class="stub"π
2
≤2x-class="stub"π
6
≤2kπ+class="stub"π
2
(k∈Z)

kπ-class="stub"π
6
≤2x-class="stub"π
3
(k∈Z).

所以f(x)的单调增区间是[kπ-class="stub"π
6
,kπ+class="stub"π
3
](k∈Z).

(Ⅲ)因为0≤x≤class="stub"π
2
,所以-class="stub"π
6
≤2x-class="stub"π
6
≤class="stub"5π
6
.

所以-class="stub"1
2
≤sin(2x-class="stub"π
6
)≤1.

所以f(x)=2sin(2x-class="stub"π
6
)+2∈[1,4].

即f(x)的最小值为1,最大值为4.

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