已知函数f(x)=cos(2x-π3)+2sin(x-π4)sin(x+π4),x∈R(I)求函数f(x)的单调递增区间与对称轴方程;(II)当x∈[-π12,π2]时,求函数f(x)的值域.-数学

题目简介

已知函数f(x)=cos(2x-π3)+2sin(x-π4)sin(x+π4),x∈R(I)求函数f(x)的单调递增区间与对称轴方程;(II)当x∈[-π12,π2]时,求函数f(x)的值域.-数学

题目详情

已知函数f(x)=cos(2x-
π
3
)+2sin(x-
π
4
)sin(x+
π
4
),x∈R
(I)求函数f(x)的单调递增区间与对称轴方程;
(II)当x∈[-
π
12
π
2
]时,求函数f(x)的值域.
题型:解答题难度:中档来源:不详

答案

(1)∵f(x)=cos(2x-class="stub"π
3
)+2sin(x-class="stub"π
4
)sin(x+class="stub"π
4

=class="stub"1
2
sin2x+
3
2
sin2x+(sinx-cosx)(sinx+cosx).
=class="stub"1
2
cos2x+
3
2
sin2x+sin2x-cos2x
=class="stub"1
2
cos2x+
3
2
sin2x-cos2x=sin(2x-class="stub"π
6

由2kπ-class="stub"π
2
≤2x-class="stub"π
6
≤2kπ+class="stub"π
2
,k∈Z,得2kπ-class="stub"π
3
≤2x≤2kπ+class="stub"2π
3
,k∈Z
kπ-class="stub"π
6
≤x≤kπ+class="stub"π
3
,k∈Z,∴单调递增区间为:[kπ-class="stub"π
6
kπ+class="stub"π
3
],k∈Z
由2x-class="stub"π
6
=kπ+class="stub"π
2
,k∈Z,得:x=class="stub"kπ
2
+class="stub"π
3
,k∈Z,
对称轴方程为x=class="stub"kπ
2
+class="stub"π
3
,k∈Z,
(2)∵x∈[-class="stub"π
12
class="stub"π
2
],∴2x-class="stub"π
6
∈[-class="stub"π
3
class="stub"5π
6
],因为f(x)=sin(2x-class="stub"π
6

在区间[-class="stub"π
12
class="stub"π
3
]上单调递增.在区间[class="stub"π
3
class="stub"π
2
]单调递减,所以当x=class="stub"π
3
,f(x)取最大值l.
又∵f(-class="stub"π
12
)=-
3
2
<f(class="stub"π
2
)=class="stub"1
2
,当x=-class="stub"π
12
时,f(x)取最小值-
3
2

所以函数f(x)在区间上的值域为[-
3
2
,1].

更多内容推荐