已知函数f(x)=cos(2x+π3)+2cos2x.(Ⅰ)求函数f(x)的周期及递增区间;(Ⅱ)求函数f(x)在[-π3,π3]上值域.-数学

题目简介

已知函数f(x)=cos(2x+π3)+2cos2x.(Ⅰ)求函数f(x)的周期及递增区间;(Ⅱ)求函数f(x)在[-π3,π3]上值域.-数学

题目详情

已知函数f(x)=cos(2x+
π
3
)+2cos2x

(Ⅰ)求函数f(x)的周期及递增区间;
(Ⅱ)求函数f(x)在[-
π
3
π
3
]
上值域.
题型:解答题难度:中档来源:不详

答案

函数f(x)=cos(2x+class="stub"π
3
)+2cos2x

=cos2xcosclass="stub"π
3
-sin2xsinclass="stub"π
3
+cos2x+1
=class="stub"3
2
cos2x-
3
2
sin2x+1

=1-
3
sin(2x-class="stub"π
3
).
(Ⅰ)函数f(x)的周期T=π,
由∵class="stub"π
2
+2kπ≤2x-class="stub"π
3
class="stub"3π
2
+2kπ,k∈Z
class="stub"5π
12
+kπ≤x≤class="stub"11π
12
+kπ,k∈Z
所以y=1-
3
sin(2x-class="stub"π
3
)的单调增区间是[class="stub"5π
12
+kπ,class="stub"11π
12
+kπ];
(Ⅱ)∵x∈[-class="stub"π
3
,class="stub"π
3
]
,∴2x-class="stub"π
3
∈[-π,class="stub"π
3
],
∴-1≤sin(2x-class="stub"π
3
)≤
3
2

函数f(x)在[-class="stub"π
3
,class="stub"π
3
]
上值域为:[-class="stub"1
2
3
+1]

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