解:(Ⅰ)∵f(x)=x
2+2ax+b=(x+a)
2-(a
2-b)
∴①當(dāng)a
2-b≤0時(shí),單調(diào)區(qū)間為:(-∞,-a]上為減,[-a,+∞)上為增;(2分)
②當(dāng)a
2-b>0時(shí),單調(diào)區(qū)間為:
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減,
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增,
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減,
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增(5分)
(Ⅱ)因?yàn)椋喝舸嬖趯?shí)數(shù)m,使得
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同時(shí)成立,即為兩變量對(duì)應(yīng)的函數(shù)值都小于等于
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的兩變量之間間隔不超過1,故須對(duì)a
2-b和
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,
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的大小分情況討論
①當(dāng)
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時(shí),由方程
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,解得
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,
此時(shí)
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,不滿足.(8分)
②當(dāng)
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時(shí),由方程
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,解得
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此時(shí)
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,滿足題意.(11分)
③當(dāng)
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時(shí),由方程
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,方程
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和解得
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,
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此時(shí)由于
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,
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所以只要
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即可,此時(shí)

,綜上所述t的最大值為
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.(16分)
分析:(Ⅰ)f(x)=(x+a)
2-a
2+b開口向上,但a
2-b的正負(fù)不定,所以在取絕對(duì)值時(shí)要分類討論.在每一種情況下分別求|f(x)|的單調(diào)區(qū)間.
(Ⅱ)存在實(shí)數(shù)m,使得
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同時(shí)成立,即為兩變量對(duì)應(yīng)的函數(shù)值都小于等于

的兩變量之間間隔不超過1,故須對(duì)a
2-b和
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,
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的大小分情況討論,求出a
2-b的取值范圍,進(jìn)而求得t的最大值.
點(diǎn)評(píng):本題考查了數(shù)學(xué)上的分類討論思想.分類討論目的是,分解問題難度,化整為零,各個(gè)擊破.