已知函数f(x)=cos(2x-π3)+sin2x-cos2x.(Ⅰ)求函数f(x)的最小正周期及图象的对称轴方程;(Ⅱ)设函数g(x)=[f(x)]2+f(x),求g(x)的值域.-数学

题目简介

已知函数f(x)=cos(2x-π3)+sin2x-cos2x.(Ⅰ)求函数f(x)的最小正周期及图象的对称轴方程;(Ⅱ)设函数g(x)=[f(x)]2+f(x),求g(x)的值域.-数学

题目详情

已知函数f(x)=cos(2x-
π
3
)+sin2x-cos2x

(Ⅰ)求函数f(x)的最小正周期及图象的对称轴方程;
(Ⅱ)设函数g(x)=[f(x)]2+f(x),求g(x)的值域.
题型:解答题难度:中档来源:昌平区一模

答案

(Ⅰ)f(x)=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
)

∴周期T=class="stub"2π
2
=π,
2x-class="stub"π
6
=kπ+class="stub"π
2
(k∈Z),得x=class="stub"kπ
2
+class="stub"π
3
(k∈Z)

∴函数图象的对称轴方程为x=class="stub"kπ
2
+class="stub"π
3
(k∈Z)

(Ⅱ)g(x)=[f(x)]2+f(x)
=sin2(2x-class="stub"π
6
)+sin(2x-class="stub"π
6
)

=[sin(2x-class="stub"π
6
)+class="stub"1
2
]2-class="stub"1
4

sin(2x-class="stub"π
6
)=-class="stub"1
2
时,g(x)取得最小值-class="stub"1
4

sin(2x-class="stub"π
6
)=1
时,g(x)取得最大值2,
所以g(x)的值域为[-class="stub"1
4
, 2]

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