已知函数f(x)=sinx3cosx3+3cos2x3.,(Ⅰ)将f(x)写成Asin(ωx+φ)的形式,并求其图象对称中心的横坐标;(Ⅱ)如果△ABC的三边a、b、c满足b2=ac,且边b所对的角为

题目简介

已知函数f(x)=sinx3cosx3+3cos2x3.,(Ⅰ)将f(x)写成Asin(ωx+φ)的形式,并求其图象对称中心的横坐标;(Ⅱ)如果△ABC的三边a、b、c满足b2=ac,且边b所对的角为

题目详情

已知函数f(x)=sin
x
3
cos
x
3
+
3
cos2
x
3
.,
(Ⅰ)将f(x)写成Asin(ωx+φ)的形式,并求其图象对称中心的横坐标;
(Ⅱ)如果△ABC的三边a、b、c满足b2=ac,且边b所对的角为x,试求x的范围及此时函数f(x)的值域.
题型:解答题难度:中档来源:奉贤区二模

答案

f(x)=class="stub"1
2
sinclass="stub"2x
3
+
3
2
(1+cosclass="stub"2x
3
)=class="stub"1
2
sinclass="stub"2x
3
+
3
2
cosclass="stub"2x
3
+
3
2
=sin(class="stub"2x
3
+class="stub"π
3
)+
3
2

(Ⅰ)由sin(class="stub"2x
3
+class="stub"π
3
)
=0
class="stub"2x
3
+class="stub"π
3
=kπ(k∈z)得x=class="stub"3k-1
2
π
,k∈z,
即对称中心的横坐标为class="stub"3k-1
2
π,k∈z;
(Ⅱ)由已知b2=ac,cosx=
a2+c2-b2
2ac
=
a2+c2-ac
2ac
≥class="stub"2ac-ac
2ac
=class="stub"1
2

class="stub"1
2
≤cosx<1,  0<x≤class="stub"π
3
,  class="stub"π
3
<class="stub"2x
3
+class="stub"π
3
≤class="stub"5π
9

|class="stub"π
3
-class="stub"π
2
|>|class="stub"5π
9
-class="stub"π
2
|
,∴sinclass="stub"π
3
<sin(class="stub"2x
3
+class="stub"π
3
)≤1
,∴
3
<sin(class="stub"2x
3
+class="stub"π
3
)≤1+
3
2

即f(x)的值域为(
3
,1+
3
2
]

综上所述,x∈(0,class="stub"π
3
]
,f(x)值域为(
3
,1+
3
2
]

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