题目内容
已知定义在R上的函数f(x) 满足条件:(1)f(x)+f(-x)=2;(2)对非零实数x,都有2f(x)+f(
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(1)求函数f(x)的解析式;
(2)设函数g(x)=
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(其中n∈N*),设 an=|AnBn|,sn为数列an 的前n项和.求证:当n≥2 时,总有 Sn2>2(
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【答案】分析:(1)令解析式中的x用
代入,得到一个方程组,消去f(
)可求出函数的解析式;
(2)先求出An的坐标,依题意得Bn与An关于y=x对称求出Bn的坐标,求出an,从而求出Sn,将sn2-sn-12进行累加可求证得结论.
解答:解:(1)由②知x≠0时
⇒
⇒f(x)=x+1(4分)
(2)g(x)=
由
⇒An(
,
)
依题意得Bn与An关于y=x对称⇒Bn(
,
)=
(6分)
⇒sn=
+
⇒sn-sn-1=
=
=sn2-
+
(8分)
⇒sn2-sn-12=
(n≥2)累加得sn2-s12=2(
++
)-(
++
)
⇒sn2=2(
++
)+1-(
++
)(10分)
又1-(
++
)>1-(
++
)=
>0
∴sn2>
(12分)
点评:本题主要考查了函数解析式的求解,以及数列的通项公式和求和,同时考查了利用累加法证明不等式,属于难题.
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(2)先求出An的坐标,依题意得Bn与An关于y=x对称求出Bn的坐标,求出an,从而求出Sn,将sn2-sn-12进行累加可求证得结论.
解答:解:(1)由②知x≠0时
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⇒f(x)=x+1(4分)
(2)g(x)=
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依题意得Bn与An关于y=x对称⇒Bn(
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⇒sn=
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⇒sn2-sn-12=
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⇒sn2=2(
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又1-(
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∴sn2>
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点评:本题主要考查了函数解析式的求解,以及数列的通项公式和求和,同时考查了利用累加法证明不等式,属于难题.

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