问题 解答题
a
=(
3
cosωx,sinωx),
b
=(sinωx,0)
,其中ω>0,函数f(x)=(
a
+
b
)•
b
+k

(1)若f(x)图象申相邻两条对称轴间的距离不小于
π
2
,求ω的取值范围.
(2)若f(x)的最小正周期为π,且当x∈[-
π
6
π
6
]
时,f(x)的最大值是
1
2
,求f(x)的解析式.
答案

(1)∵

a
=(
3
cosωx,sinωx),
b
=(sinωx,0),

a
+
b
=(
3
cosωx+sinωx
,sinωx),

f(x)=(

a
+
b
)•
b
+k

=

3
sinωxcosωx+sin2ωx+k

=

3
2
sin2ωx+
1-cos2ωx
2
+k

=

3
2
sin2ωx-
1
2
cos2ωx+
1
2
+k

=sin(2ωx-

π
6
)+k+
1
2

∵f(x)图象中相邻两条对称轴间的距离不小于

π
2

T
2
=
π
π
2
,∴ω≤1,

∵ω>0,∴0<ω≤1.

(2)∵T=

=π,∴ω=1,

∴f(x)=sin(2x-

π
6
)+k+
1
2

∵x∈[-

π
6
π
6
],

∴2x-

π
6
∈[-
π
2
π
6
],

从而当2x-

π
6
=
π
6
,即x=
π
6
时,

f(x)max=f(

π
6
)=sin
π
6
+k+
1
2
=k+1=
1
2

∴k=-

1
2

故f(x)=sin(2x-

π
6
).

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