已知函数f(x)=sin2x+23sin(x+π4)cos(x-π4)-cos2x-3.(Ⅰ)求函数f(x)的最小正周期和单调递减区间;(Ⅱ)求函数f(x)在[-π12,2536π]上的最大值和最小值

题目简介

已知函数f(x)=sin2x+23sin(x+π4)cos(x-π4)-cos2x-3.(Ⅰ)求函数f(x)的最小正周期和单调递减区间;(Ⅱ)求函数f(x)在[-π12,2536π]上的最大值和最小值

题目详情

已知函数f(x)=sin2x+2
3
sin(x+
π
4
)cos(x-
π
4
)-cos2x-
3

(Ⅰ)求函数f(x)的最小正周期和单调递减区间;
(Ⅱ)求函数f(x)在[-
π
12
25
36
π]
上的最大值和最小值并指出此时相应的x的值.
题型:解答题难度:中档来源:不详

答案

(Ⅰ)f(x)=sin2x+2
3
sin(x+class="stub"π
4
)cos(x-class="stub"π
4
)-cos2x-
3

=2
3
sin2(x+class="stub"π
4
)-cos2x-
3
=
3
sin2x-cos2x
=2sin(2x-class="stub"π
6
)

所以T=class="stub"2π
2

2kπ+class="stub"π
2
≤2x-class="stub"π
6
≤2kπ+class="stub"3π
2
(k∈Z)
kπ+class="stub"π
3
≤x≤kπ+class="stub"5π
6
(k∈Z)

所以函数f(x)的最小正周期为π,单调递减区间为[kπ+class="stub"π
3
,kπ+class="stub"5π
6
]
(k∈Z).
(Ⅱ)由(Ⅰ)有f(x)=2sin(2x-class="stub"π
6
)

因为x∈[-class="stub"π
12
,class="stub"25
36
π]

所以2x-class="stub"π
6
∈[-class="stub"π
3
,class="stub"11
9
π]

因为sin(-class="stub"π
3
)=sinclass="stub"4
3
π<sinclass="stub"11
9
π

所以当x=-class="stub"π
12
时,函数f(x)取得最小值-
3

x=class="stub"π
3
时,函数f(x)取得最大值2.

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