题目内容
若
=(x1,y1),
=(x2,y2),定义:
•
=x1x2+y1y2,已知
=(2cosx,1),
=(cosx,
sin2x),f(x)=
•
,x∈R
(1)若f(x)=1-
,且x∈[-
,
],求x;
(2)若函数y=2sin2x的图象向左(或右)平移|m|(|m|<
)个单位,再向上(或下)平移|n|个单位后得到函数y=f(x)的图象,求实数m,n的值.
| a |
| b |
| a |
| b |
| a |
| b |
| 3 |
| a |
| b |
(1)若f(x)=1-
| 3 |
| π |
| 3 |
| π |
| 3 |
(2)若函数y=2sin2x的图象向左(或右)平移|m|(|m|<
| π |
| 2 |
(1)f(x)=2cos2x+
sin2x
=1+cos2x+
sin2x
=2sin(2x+
)+1=1-
,
∴sin(2x+
)=-
.
∵x∈[-
,
],
∴(2x+
)∈[-
,
].
∴2x+
=-
,解得x=-
.
(2)把函数y=2sin2x的图象向左平移
个单位,再向上平移1个单位即可得出f(x)=2sin2(x+
)+1=2sin(2x+
)+1.
∴m=-
,n=1.
| 3 |
=1+cos2x+
| 3 |
=2sin(2x+
| π |
| 6 |
| 3 |
∴sin(2x+
| π |
| 6 |
| ||
| 2 |
∵x∈[-
| π |
| 3 |
| π |
| 3 |
∴(2x+
| π |
| 6 |
| π |
| 2 |
| 5π |
| 6 |
∴2x+
| π |
| 6 |
| π |
| 3 |
| π |
| 4 |
(2)把函数y=2sin2x的图象向左平移
| π |
| 12 |
| π |
| 12 |
| π |
| 6 |
∴m=-
| π |
| 12 |
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若
=(x1,y1),
=(x2,y2),且
∥
,则有( )
| a |
| b |
| a |
| b |
| A、x1y2+x2y1=0 |
| B、x1y2-x2y1=0, |
| C、x1x2+y1y2=0 |
| D、x1x2-y1y2=0 |