已知函数f(x)=1-2sin2(x+π24)+2sin(x+π24)cos(x+π24).(I)求f(x)的最小正周期;(II)求函数f(x)的单调递增区间.-数学

题目简介

已知函数f(x)=1-2sin2(x+π24)+2sin(x+π24)cos(x+π24).(I)求f(x)的最小正周期;(II)求函数f(x)的单调递增区间.-数学

题目详情

已知函数f(x)=1-2sin2(x+
π
24
)+2sin(x+
π
24
)cos(x+
π
24
).
(I)求f(x)的最小正周期;
(II)求函数f(x)的单调递增区间.
题型:解答题难度:中档来源:和平区一模

答案

(Ⅰ)函数f(x)=1-2sin2(x+class="stub"π
24
)+2sin(x+class="stub"π
24
)cos(x+class="stub"π
24

=
2
cos(2x+class="stub"π
12
)•cosclass="stub"π
4
+cos(2x+class="stub"π
12
)sinclass="stub"π
4

=
2
sin(2x+class="stub"π
3

∴函数的最小正周期为:T=class="stub"2π
2
=π.
(Ⅱ)由(Ⅰ)可知f(x)=
2
sin(2x+class="stub"π
3

当-class="stub"π
2
+2kπ≤2x+class="stub"π
3
class="stub"π
2
+2kπ(k∈Z),
即kπ-class="stub"5π
12
≤x≤class="stub"π
12
+kπ,k∈Z.时函数是增函数.
所以函数的单调增区间为:[kπ-class="stub"5π
12
class="stub"π
12
+kπ],k∈Z.

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