题目内容
已知
=(sinx,cosx),
=(
cosx,cosx),设函数f(x)=
•
(x∈R)
(1)求f(x)的最小正周期及单调递增区间;
(2)当x∈[-
,
]时,求f(x)的值域.
a |
b |
3 |
a |
b |
(1)求f(x)的最小正周期及单调递增区间;
(2)当x∈[-
π |
6 |
5π |
12 |
(1)∵f(x)=
•
=
sin2x+
cos2x+
=sin(2x+
)+
,∴f(x)的最小正周期为π.
由-
+2kπ≤2x+
≤
+2kπ得,-
+kπ≤x≤
+kπ,(k∈Z),解得 -
+kπ≤x≤
+kπ,
故f(x)的单调增区间为[-
+kπ,
+kπ],(k∈Z).
(2)由(1)知f(x)=sin(2x+
)+
,又当x∈[-
,
],2x+
∈[-
,π],故-
≤sin(2x+
)≤1,
从而f(x)的值域为[0,
].
a |
b |
| ||
2 |
1 |
2 |
1 |
2 |
π |
6 |
1 |
2 |
由-
π |
2 |
π |
6 |
π |
2 |
π |
3 |
π |
6 |
π |
3 |
π |
6 |
故f(x)的单调增区间为[-
π |
3 |
π |
6 |
(2)由(1)知f(x)=sin(2x+
π |
6 |
1 |
2 |
π |
6 |
5π |
12 |
π |
6 |
π |
6 |
1 |
2 |
π |
6 |
从而f(x)的值域为[0,
3 |
2 |
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