题目内容
已知
=(cos
,sin
),
=(cos
,-sin
),且x∈[-
,
].
(Ⅰ)求
•
及|
+
|
(Ⅱ)若f(x)=
•
-|
+
|,求f(x)的最大值和最小值.
| a |
| 3x |
| 2 |
| 3x |
| 2 |
| b |
| x |
| 2 |
| x |
| 2 |
| π |
| 3 |
| π |
| 4 |
(Ⅰ)求
| a |
| b |
| a |
| b |
(Ⅱ)若f(x)=
| a |
| b |
| a |
| b |
(Ⅰ)由题意可得:因为
=(cos
,sin
),
=(cos
,-sin
),
所以
•
=cos
cos
-sin
sin
=cos2x,
所以|
+
|=
=2|cosx|=2cosx,x∈[-
,
].
(Ⅱ)由(I)可得:f(x)=
•
-|
+
|
=cos2x-2cosx
=2cos2x-1-2cosx
=2(cosx-
)2-
∵x∈[-
,
]
∴cosx∈[
,1],
设t=cosx,则t∈[
,1],
所以y=2(t-
)2-
,
∴f(x)max=-1,f(x)min=-
.
| a |
| 3x |
| 2 |
| 3x |
| 2 |
| b |
| x |
| 2 |
| x |
| 2 |
所以
| a |
| b |
| 3x |
| 2 |
| x |
| 2 |
| 3x |
| 2 |
| x |
| 2 |
所以|
| a |
| b |
|
|
| π |
| 3 |
| π |
| 4 |
(Ⅱ)由(I)可得:f(x)=
| a |
| b |
| a |
| b |
=cos2x-2cosx
=2cos2x-1-2cosx
=2(cosx-
| 1 |
| 2 |
| 3 |
| 2 |
∵x∈[-
| π |
| 3 |
| π |
| 4 |
∴cosx∈[
| ||
| 2 |
设t=cosx,则t∈[
| ||
| 2 |
所以y=2(t-
| 1 |
| 2 |
| 3 |
| 2 |
∴f(x)max=-1,f(x)min=-
| 2 |
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