题目内容
曲线y=2sin(x+
)cos(x-
)和直线y=
在y轴右侧的交点按横坐标从小到大依次记为P1,P2,P3,…,则|P2P4|等于 ( )
| π |
| 4 |
| π |
| 4 |
| 1 |
| 2 |
分析:本题考查的知识点是诱导公式,二倍角公式及函数图象的交点,将y=2sin(x+
)cos(x-
)=sin2x+1令y=
,解得x=kπ+
±
(k∈N),代入易得|P2P4|的值.
| π |
| 4 |
| π |
| 4 |
| 1 |
| 2 |
| 3π |
| 4 |
| π |
| 6 |
解答:解:∵y=2sin(x+
)cos(x-
)
=2sin(x-
+
)cos(x-
)
=2cos(x-
)cos(x-
)
=cos[2(x-
)]+1
=sin2x+1
若y=2sin(x+
)cos(x-
)=
则2x=2kπ+
±
(k∈N)
x=kπ+
±
(k∈N)
故|P2P4|=π
故选:A
| π |
| 4 |
| π |
| 4 |
=2sin(x-
| π |
| 4 |
| π |
| 2 |
| π |
| 4 |
=2cos(x-
| π |
| 4 |
| π |
| 4 |
=cos[2(x-
| π |
| 4 |
=sin2x+1
若y=2sin(x+
| π |
| 4 |
| π |
| 4 |
| 1 |
| 2 |
则2x=2kπ+
| 3π |
| 2 |
| π |
| 3 |
x=kπ+
| 3π |
| 4 |
| π |
| 6 |
故|P2P4|=π
故选:A
点评:求两个函数图象的交点间的距离,关于是要求出交点的坐标,然后根据两点间的距离求法进行求解.
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