题目内容
已知函数f(x)=msinx+ncosx,且
①
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②函数f(x)的图象关于点
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③
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④记函数f(x)的图象在y轴右侧与直线
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⑤
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其中真命题的是 (写出所有正确命题的编号)
【答案】分析:由题意可得f(x)=
sin(x+
),对于①,由于
=cosx,是偶函数,故①正确.
对于②,由于当x=
时,f(x)=0,故②正确.
对于③,由于
=-
,是 函数f(x)的最小值,故 ③正确.
对于④,由题意可得,|P2P4|等于一个周期2π,故 ④不正确.
对于⑤,由tan∅=tan(2kπ+
)=
=1,可得⑤正确.
解答:解:由于函数f(x)=msinx+ncosx=
sin(x+∅),且
是它的最大值,
∴
+∅=2kπ+
,k∈z,∴∅=2kπ+
,∴tan∅=
=1.
∴f(x)=
sin(x+2kπ+
)=
sin(x+
).
对于①,由于
=
sin(x+
+
)=cosx,是偶函数,故①正确.
对于②,由于当x=
时,f(x)=0,故函数f(x)的图象关于点
对称,故②正确.
对于③,由于
=
sin(-
)=-
,是 函数f(x)的最小值,故 ③正确.
对于④,函数f(x)的图象即把函数 y=
sinx的图象向左平移
个单位得到的,故|P2P4|等于
一个周期2π,故 ④不正确.
对于⑤,由tan∅=
=1,可得⑤正确.
故答案为:①②③⑤.
点评:本题考查两角和正弦公式,正弦函数的最值,对称性,奇偶性,函数图象的变换,得到 f(x)=
sin(x+
),是解题的关键.
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对于②,由于当x=
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对于③,由于
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对于④,由题意可得,|P2P4|等于一个周期2π,故 ④不正确.
对于⑤,由tan∅=tan(2kπ+

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解答:解:由于函数f(x)=msinx+ncosx=

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∴
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∴f(x)=
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对于①,由于
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对于②,由于当x=

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对于③,由于

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对于④,函数f(x)的图象即把函数 y=
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
一个周期2π,故 ④不正确.
对于⑤,由tan∅=

故答案为:①②③⑤.
点评:本题考查两角和正弦公式,正弦函数的最值,对称性,奇偶性,函数图象的变换,得到 f(x)=

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