题目内容
已知向量
=(sinx,-1),
=(
cosx,-
),函数f(x)=(
+
)•
-2.
(Ⅰ)求函数f(x)的最小正周期T:
(Ⅱ)若x∈[
,
],试求f(x)的取值范围.
| a |
| b |
| 3 |
| 1 |
| 2 |
| a |
| b |
| a |
(Ⅰ)求函数f(x)的最小正周期T:
(Ⅱ)若x∈[
| π |
| 6 |
| 5π |
| 6 |
(Ⅰ)由题意可得,函数f(x)=(
+
)•
-2=
2+
•
-2=1+sin2x+
sinxcosx+
-2
=
sin2x+
-
=sin(2x-
),
故函数f(x)的最小正周期T=
=π.
(Ⅱ)若x∈[
,
],
≤2x-
≤
,故当2x-
=
时,f(x)取得最小值为-1,
当2x-
=
时,f(x)取得最大值为1,
故函数f(x)的取值范围是[-1,1].
| a |
| b |
| a |
| a |
| a |
| b |
| 3 |
| 1 |
| 2 |
=
| ||
| 2 |
| 1-cos2x |
| 2 |
| 1 |
| 2 |
| π |
| 6 |
故函数f(x)的最小正周期T=
| 2π |
| 2 |
(Ⅱ)若x∈[
| π |
| 6 |
| 5π |
| 6 |
| π |
| 6 |
| π |
| 6 |
| 3π |
| 2 |
| π |
| 6 |
| 3π |
| 2 |
当2x-
| π |
| 6 |
| π |
| 2 |
故函数f(x)的取值范围是[-1,1].
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