题目内容
已知向量
=(cos
x,sin
x),
=(cos
-sin
),x∈[0,
]
(1)用x的式子表示;
.
及|
+
|;
(2)求函数f(x)=
.
-4|
+
|的值域;
(3)设g(x)=
.
+t|
+
|,若关于x的方程g(x)+2=0有两不同解,求t的取值范围?.
| a |
| 3 |
| 2 |
| 3 |
| 2 |
| b |
| x |
| 2 |
| x |
| 2 |
| π |
| 2 |
(1)用x的式子表示;
| a |
| b |
| a |
| b |
(2)求函数f(x)=
| a |
| b |
| a |
| b |
(3)设g(x)=
| a |
| b |
| a |
| b |
(1)
•
=cos
cos
-sin
sin
=cos2x
∵|
+
|2=1+2cos2x+1=2(1+cos2x)=4cos2x
∴|
+
|=2cosx x∈[0,
]
(2)∵f(x)=
•
-4|
+
|=cos2x-8cosx=2cos2x-8cosx-1=2(cosx-2)2-9
∵x∈[0,
]∴cosx∈[0,1]∴f(x)∈[-7,-1]
(3)∵g(x)+2=0
∴cos2x+2tcosx+2=0
即2cos2x+2tcosx+1=0
令cosx=μ∈[0,1),F(μ)=2μ2+2tμ+1
∴
∴t∈[-
,-
)
| a |
| b |
| 3x |
| 2 |
| x |
| 2 |
| 3x |
| 2 |
| x |
| 2 |
∵|
| a |
| b |
∴|
| a |
| b |
| π |
| 2 |
(2)∵f(x)=
| a |
| b |
| a |
| b |
∵x∈[0,
| π |
| 2 |
(3)∵g(x)+2=0
∴cos2x+2tcosx+2=0
即2cos2x+2tcosx+1=0
令cosx=μ∈[0,1),F(μ)=2μ2+2tμ+1
∴
|
∴t∈[-
| 3 |
| 2 |
| 2 |
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