题目内容
已知向量
=(cos
,sin
),
=(cos
,-sin
),且x∈[
,π].
(1)求
•
及|
+
|;
(2)求函数f(x)=
•
+|
+
|的最大值,并求使函数取得最大值时x的值.
| a |
| 3x |
| 2 |
| 3x |
| 2 |
| b |
| x |
| 2 |
| x |
| 2 |
| π |
| 2 |
(1)求
| a |
| b |
| a |
| b |
(2)求函数f(x)=
| a |
| b |
| a |
| b |
(1)∵向量
=(cos
,sin
),
=(cos
,-sin
),且x∈[
,π].
∴
•
=cos
cos
-sin
sin
=cos2x,
|
+
| =
=
=
=2|cosx|,
∵x∈[
,π],
∴cosx<0.
∴|
+
|=-2cosx.
(2)f(x)=
•
+|
+
|
=cos2x-2cosx
=2cos2x-2cosx-1
=2(cosx-
)2-
,
∵x∈[
,π],
∴-1≤cosx≤0,…(13分)
∴当cosx=-1,即x=π时,fmax(x)=3.
| a |
| 3x |
| 2 |
| 3x |
| 2 |
| b |
| x |
| 2 |
| x |
| 2 |
| π |
| 2 |
∴
| a |
| b |
| 3x |
| 2 |
| x |
| 2 |
| 3x |
| 2 |
| x |
| 2 |
=cos2x,
|
| a |
| b |
(cos
|
=
2+2(cos
|
=
| 2+2cos2x |
=2|cosx|,
∵x∈[
| π |
| 2 |
∴cosx<0.
∴|
| a |
| b |
(2)f(x)=
| a |
| b |
| a |
| b |
=cos2x-2cosx
=2cos2x-2cosx-1
=2(cosx-
| 1 |
| 2 |
| 3 |
| 2 |
∵x∈[
| π |
| 2 |
∴-1≤cosx≤0,…(13分)
∴当cosx=-1,即x=π时,fmax(x)=3.
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