已知函数f(x)=2-2cosx+2-2cos(2π3-x),x∈[0,2π],则当x=______时,函数f(x)有最大值,最大值为______.-数学

题目简介

已知函数f(x)=2-2cosx+2-2cos(2π3-x),x∈[0,2π],则当x=______时,函数f(x)有最大值,最大值为______.-数学

题目详情

已知函数f(x)=
2-2cosx
+
2-2cos(
3
-x)
,x∈[0,2π],则当x=______时,函数f(x)有最大值,最大值为______.
题型:填空题难度:中档来源:不详

答案

∵函数f(x)=
2-2cosx
+
2-2cos(class="stub"2π
3
-x)

=
2(1-cosx)
+
2[1-cos(x-class="stub"2π
3
)

=2|sinclass="stub"x
2
|+2|sin(class="stub"x
2
-class="stub"π
3
)|
=
2sinclass="stub"x
2
-2sin(class="stub"x
2
-class="stub"π
3
)      0≤x≤class="stub"2π
3
2sinclass="stub"x
2
+2sin(class="stub"x
2
-class="stub"π
3
)      class="stub"2π
3
<x≤2π

=
2sin(class="stub"x
2
+class="stub"π
3
)   0≤x≤class="stub"2π
3
2
3
sin(class="stub"x
2
-class="stub"π
6
)   class="stub"2π
3
<x≤2π  

class="stub"x
2
+class="stub"π
3
=class="stub"π
2
⇒x=class="stub"π
3
时,y=2sin(class="stub"x
2
+class="stub"π
3
)有最大值2;
class="stub"x
2
-class="stub"π
6
=class="stub"π
2
⇒x=class="stub"4π
3
时,y=2
3
sin(class="stub"x
2
-class="stub"π
6
)有最大值2
3

∴当x=class="stub"4π
3
时,函数f(x)有最大值2
3

故答案为:class="stub"4π
3
,2
3

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