题目内容
【题目】已知函数f(x)=2cos(x+ )[sin(x+
)﹣
cos(x+
)].
(1)求f(x)的值域和最小正周期;
(2)若对任意x∈[0, ],[f(x)+
]﹣2m=0成立,求实数m的取值范围.
【答案】
(1)解:函数f(x)=2cos(x+ )[sin(x+
)﹣
cos(x+
)]
=2cos(x+ )sin(x+
)﹣2
cos2(x+
)
=sin(2x+ )﹣2
=sin(2x+ )﹣
cos(2x+
)﹣
=2sin[(2x+ )﹣
]﹣
=2sin(2x+ )﹣
,
∴函数f(x)的最小正周期为T= =
=π;
又﹣1≤sin(2x+ )≤1,
∴﹣2﹣ ≤2sin(2x+
)﹣
≤2﹣
,
即f(x)的值域为[﹣2﹣ ,2﹣
];
(2)解:对任意x∈[0, ],[f(x)+
]﹣2m=0成立,
∴[2sin(2x+ )﹣
+
]﹣2m=0,
即sin(2x+ )=m;
由x∈[0, ],得2x+
∈[
,
],
∴sin(2x+ )∈[
,1],
∴实数m的取值范围是m∈[ ,1].
【解析】(1)化简函数f(x)为正弦型函数,求出它的最小正周期和值域;(2)对任意x∈[0, ],[f(x)+
]﹣2m=0成立,等价于sin(2x+
)=m;求出x∈[0,
]时sin(2x+
)的值域即可.
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