题目内容
已知向量
=(cos
x,sin
x),
=(cos
,-sin
),
=(-sin
,cos
),且x∈[-
,
].
(1)求|
+
|;
(2)求函数f(x)=2
•
+|
+
|的单调增区间.
| a |
| 3 |
| 2 |
| 3 |
| 2 |
| b |
| x |
| 2 |
| x |
| 2 |
| c |
| x |
| 2 |
| x |
| 2 |
| π |
| 2 |
| π |
| 2 |
(1)求|
| a |
| b |
(2)求函数f(x)=2
| a |
| c |
| a |
| b |
分析:(1)根据
=(cos
x,sin
x),
=(cos
,-sin
),可得|
+
|2=
2+2
•
+
2,利用x∈[-
,
],即可求得|
+
|;
(2)函数f(x)=2
•
+|
+
|=2sinx+2cosx=2
sin(x+
),x∈[-
,
],令μ=x+
,则可得μ的范围,y=sinμ在[-
,
]上为增函数,由此可得函数f(x)=2
•
+|
+
|单调增区间.
| a |
| 3 |
| 2 |
| 3 |
| 2 |
| b |
| x |
| 2 |
| x |
| 2 |
| a |
| b |
| a |
| a |
| b |
| b |
| π |
| 2 |
| π |
| 2 |
| a |
| b |
(2)函数f(x)=2
| a |
| c |
| a |
| b |
| 2 |
| π |
| 4 |
| π |
| 2 |
| π |
| 2 |
| π |
| 4 |
| π |
| 4 |
| π |
| 2 |
| a |
| c |
| a |
| b |
解答:解:(1)∵
=(cos
x,sin
x),
=(cos
,-sin
)
∴|
+
|2=
2+2
•
+
2=2+2cos2x=4cos2x
∵x∈[-
,
]
∴cosx>0
∴|
+
|=2cosx;
(2)
•
=sin(
x-
)=sinx
∴f(x)=2
•
+|
+
|=2sinx+2cosx=2
sin(x+
)
其中x∈[-
,
],令μ=x+
,则μ∈[-
,
],y=sinμ在[-
,
]上为增函数
由μ∈[-
,
]可得x∈[-
,
],故sin(x+
)的增区间为[-
,
]
即函数f(x)=2
•
+|
+
|单调增区间为[-
,
]
| a |
| 3 |
| 2 |
| 3 |
| 2 |
| b |
| x |
| 2 |
| x |
| 2 |
∴|
| a |
| b |
| a |
| a |
| b |
| b |
∵x∈[-
| π |
| 2 |
| π |
| 2 |
∴cosx>0
∴|
| a |
| b |
(2)
| a |
| c |
| 3 |
| 2 |
| x |
| 2 |
∴f(x)=2
| a |
| c |
| a |
| b |
| 2 |
| π |
| 4 |
其中x∈[-
| π |
| 2 |
| π |
| 2 |
| π |
| 4 |
| π |
| 4 |
| 3π |
| 4 |
| π |
| 4 |
| π |
| 2 |
由μ∈[-
| π |
| 4 |
| π |
| 2 |
| π |
| 2 |
| π |
| 4 |
| π |
| 4 |
| π |
| 2 |
| π |
| 4 |
即函数f(x)=2
| a |
| c |
| a |
| b |
| π |
| 2 |
| π |
| 4 |
点评:本题考查向量知识的综合运用,考查向量的模,考查三角函数的单调性,解题的关键是利用三角函数知识求解.
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