試題分析:(1)本題唯一的條件是
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為奇函數(shù),故其定義域關(guān)于原點(diǎn)對稱,通過求函數(shù)的定義域可求得
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,當(dāng)然這時(shí)還要根據(jù)奇函數(shù)的定義驗(yàn)證
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確實(shí)是奇函數(shù);(2)要判斷函數(shù)的單調(diào)性,可根據(jù)復(fù)合函數(shù)單調(diào)性的性質(zhì)確定,然后再根據(jù)定義證明,而函數(shù)
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為奇函數(shù),故只要判斷函數(shù)在區(qū)間
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上的單調(diào)性即可,變形
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為
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可得
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在
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是遞減,當(dāng)然它在
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上也是遞減的,然后用單調(diào)性定義田加以證明;(3)
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為奇函數(shù),它的對稱中心為
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,
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的圖象是由
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的圖象平移過去的,因此對稱中心也相應(yīng)平移,即
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對稱中心為
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,函數(shù)
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的圖象對稱中心為
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,則
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有性質(zhì):
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,因此本題是有
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,即
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.
試題解析:(1)因?yàn)楹瘮?shù)為奇函數(shù),所以定義域關(guān)于原點(diǎn)對稱,由
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,得
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,所以
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. 2分
這時(shí)
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滿足
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,函數(shù)為奇函數(shù),因此
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4分
(2)函數(shù)為單調(diào)遞減函數(shù).
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法一:用單調(diào)性定義證明;
法二:利用已有函數(shù)的單調(diào)性加以說明.
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在
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上單調(diào)遞增,因此
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單調(diào)遞增,又
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在
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及
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上單調(diào)遞減,因此函數(shù)
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在
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及
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上單調(diào)遞減;
法三:函數(shù)定義域?yàn)?img src="http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824032205444631.png" style="vertical-align:middle;" />,說明函數(shù)在
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上單調(diào)遞減,因?yàn)楹瘮?shù)為奇函數(shù),因此函數(shù)在
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上也是單調(diào)遞減,因此函數(shù)
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在
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及
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上單調(diào)遞減.
10分
(本題根據(jù)具體情況對照給分)
(3)因?yàn)楹瘮?shù)
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為奇函數(shù),因此其圖像關(guān)于坐標(biāo)原點(diǎn)(0,0)對稱,根據(jù)條件得到函數(shù)
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的一個(gè)對稱中心為
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, 13分
因此有
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,因?yàn)?img src="http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824032204695511.png" style="vertical-align:middle;" />,因此
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16分