已知向量a=(sinx3,cosx3),b=(cosx3,3cosx3),函数f(x)=a•b,(1)求函数f(x)的单调递增区间;(2)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

已知向量a=(sinx3,cosx3),b=(cosx3,3cosx3),函数f(x)=a•b,(1)求函数f(x)的单调递增区间;(2)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

已知向量
a
=(sin
x
3
,cos
x
3
),
b
=(cos
x
3
3
cos
x
3
),函数f(x)=
a
b

(1)求函数f(x)的单调递增区间;
(2)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:不详

答案

(1)∵向量
a
=(sinclass="stub"x
3
,cosclass="stub"x
3
b
=(cosclass="stub"x
3
3
cosclass="stub"x
3
),
∴函数f(x)=
a
b
=sin(class="stub"2x
3
+class="stub"π
3
)+
3
2

令2kπ-class="stub"π
2
class="stub"2x
3
+class="stub"π
3
≤2kπ+class="stub"π
2
,解得3kπ-class="stub"5
4
π≤x≤3kπ+class="stub"π
4
(k∈Z)

故函数f(x)的单调递增区间为[3kπ-class="stub"5
4
π,3kπ+class="stub"π
4
](k∈Z)

(2)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

3
2
<sin(class="stub"2x
3
+class="stub"π
3
)≤1,
3
<sin(class="stub"2x
3
+class="stub"π
3
)+
3
2
≤1+
3
2

∴f(x)的值域为(
3
,1+
3
2
]

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