已知函数f(x)=2cosxsin(x+π3)-3sin2x+sinxcosx(1)求f(x)的最小正周期;(2)求f(x)的单调增区间;(3)当x∈[0,π4]时,求f(x)的值域.-数学

题目简介

已知函数f(x)=2cosxsin(x+π3)-3sin2x+sinxcosx(1)求f(x)的最小正周期;(2)求f(x)的单调增区间;(3)当x∈[0,π4]时,求f(x)的值域.-数学

题目详情

已知函数f(x)=2cosxsin(x+
π
3
)-
3
sin2x+sinxcosx
(1)求f(x)的最小正周期;
(2)求f(x)的单调增区间;
(3)当x∈[0,
π
4
]
时,求f(x)的值域.
题型:解答题难度:中档来源:苏州一模

答案

f(x)=2cosxsin(x+class="stub"π
3
)-
3
(sinx)2+sinxcosx=2cosx(sinclass="stub"x
2
+
3
cosclass="stub"x
2
)-
3
class="stub"1-cos2x
2
+class="stub"1
2
sin2x
=sinxcosx+
3
class="stub"1-cosx
2
-
3
2
+
3
class="stub"cos2x
2
+class="stub"sin2x
2

=sin2x+
3
cos2x
=2sin(2x+class="stub"π
3

(1)因为T=class="stub"2π
|ω|
=class="stub"2π
2
=π,所以函数的最小正周期是π.
(2)y=sinx的单调增区间是[2kπ-class="stub"π
2
,2kπ+class="stub"π
2
]k∈Z,则函数f(x)=2cosxsin(x+class="stub"π
3
)-
3
sin2x+sinxcosx
即:2sin(2x+class="stub"π
3
)的单增区间:2x+class="stub"π
3
∈[2kπ-class="stub"π
2
,2kπ+class="stub"π
2
]
解得x∈[kπ-class="stub"5π
12
,kπ+class="stub"π
12
](k∈Z)
(3)x∈[0,class="stub"π
4
]
,则2x+class="stub"π
3
∈[class="stub"π
3
class="stub"5π
6
],所以2sin(2x+class="stub"π
3
)∈[class="stub"1
2
,1]
所以函数的值域为:[class="stub"1
2
,1].

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