解:(1)設(shè)g(x)=4x
2-x-b(x≥
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)
令g′(x)=8x-1=0,可得x=
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,
∵
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,∴g(x)在[
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,+∞)上單調(diào)增;
g(x)=-2x
2+x-b(x<
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)
令g′(x)=-4x+1=0,可得x=
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,
∵
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,∴g(x)在(-∞,
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)上單調(diào)增;g(x)在[
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,
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)上單調(diào)減;
要使方程f(x)=b恰有三個根,只須g(
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)=-2(
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)
2+
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-b=
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-b>0,∴b<
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g(
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)=-2(
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)
2+
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-b=
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-b<0,∴b>
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∴
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;
(2)當(dāng)m<n≤
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時,f(x)在區(qū)間[m,n]上單調(diào)遞增,所以
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,所以m=n,矛盾;
當(dāng)m≤
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≤n<
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時,n=f(
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)=
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,矛盾;
當(dāng)m≤
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<
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≤n時,n≥
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>
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>f(m),故f(x)在區(qū)間[m,n]上的最大值在[
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,n]上取到
∵f(x)在[
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,n]上單調(diào)遞增,∴n=f(n),∴n=
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又
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,故
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,所以f(x)在區(qū)間[m,n]上的最小值在
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上取到.
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又f(x)在區(qū)間
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上單調(diào)遞增,故m=f(m),∴m=0
故
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當(dāng)
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時,由x∈
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,
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知,
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,矛盾.
當(dāng)
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時,f(x)在區(qū)間
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上單調(diào)遞減,
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上單調(diào)遞增.故
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,矛盾
當(dāng)
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時,f(x)在區(qū)間[m,n]上單調(diào)遞增,故
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,得
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,矛盾.
綜上所述
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,即存在區(qū)間
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滿足條件.
(3)當(dāng)a>0時,函數(shù)的圖象如右,
要使得函數(shù)f(x)在開區(qū)間(m,n)內(nèi)既有最大值又有最小值,則最小值一定在x=a處取得,最大值在
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處取得;
f(a)=a
2,在區(qū)間(-∞,a)內(nèi),函數(shù)值為a
2時
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,所以
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;
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,而在區(qū)間(a,+∞)內(nèi)函數(shù)值為
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時
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,所以
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.…..(12分)
分析:(1)利用絕對值的幾何意義,分類討論,確定函數(shù)的單調(diào)性,從而要使方程f(x)=b恰有三個根,只須g(
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)>0,g(
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)<0,從而可求實數(shù)b的取值范圍;
(2)分類討論,確定函數(shù)的單調(diào)性,求出函數(shù)的最值,即可求得結(jié)論;
(3)要使函數(shù)在(m,n)上既有最大值又有最小值,則最小值在x=a處取得,最大值在
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處取得.
點評:本題考查函數(shù)的最值,考查分類討論的數(shù)學(xué)思想,考查學(xué)生分析解決問題的能力,屬于中檔題.