函數(shù)f(x)=x4-2ax2,g(x)=1.
(1)求證:函數(shù)f(x)與g(x)的圖象恒有公共點;
(2)當x∈(0,1]時,若函數(shù)f(x)圖象上任一點處切線斜率均小于1,求實數(shù)a的取值范圍;
(3)當x∈[0,1]時,關于x的不等式|f′(x)|>g(x)的解集為空集,求所有滿足條件的實數(shù)a的值.
【答案】
分析:(1)兩個函數(shù)的交點轉(zhuǎn)化為一個函數(shù)與x軸的交點,轉(zhuǎn)化為對應方程的有實數(shù)解,換元轉(zhuǎn)化為二次方程有非負實數(shù)根,由送別式恒大于0與兩根之積為負得二次方程一定有正根,問題得證.
(2)求導,由題意得導數(shù)恒小于1,分離參數(shù)a,設另一邊為函數(shù),求導得導數(shù)恒大于0,函數(shù)在(0,1]上遞增,得最值,求出參數(shù)a的取值范圍;
(3)把函數(shù)解析式代入不等式,考慮反面,轉(zhuǎn)化為恒成立問題,設絕對值符號內(nèi)的為F(x),求導,得函數(shù)單調(diào)性,結(jié)合函數(shù)圖象,討論函數(shù)在[0,1]上的單調(diào)性,進而求出最值,令最值的絕對值小于等于1,得實數(shù)a的值.
解答:解:(1)設h(x)=f(x)-g(x)
即證函數(shù)h(x)與x軸有交點,
即證方程x
4-2ax
2-1=0有實根,設t=x
2即證方程t
2-2at-1=0有非負實數(shù)根,
而△=4a
2+4>0,t
1t
2=-1<0
∴方程t
4-2at-1=0恒有正根
∴f(x)與g(x)圖象恒有公共點(4分)
(2)f′(x)=4x
3-4ax
∵當0<x≤1時4xa>4x
3-1恒成立
即
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,設y=x
2-
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,
則y′=2x+
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>0,
∴y=x
2-
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在(0,1]上單調(diào)遞增,
∴a>1-
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=
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∴a的取值范圍為
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(8分)
(3)由題設知當x∈[0,1]時,|4x
3-4ax|≤1恒成立
記F(x)=4x
3-4ax
若a≤0則F(1)=4(1-a)≥4不滿足條件
故a>0而
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①當
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時,即0<a<3時,F(xiàn)(x)在
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上遞減,在
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上遞增,
于是
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∴
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,∴
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,∴
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②當
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時,即a≥3時,F(xiàn)(x)在[0,1]上遞減,
于是
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矛盾
綜上所述:
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(14分)
點評:本題考查導數(shù)在最大值、最小值問題中的應用,一是當問題從正面不容易解決時,注意從反面進行突破,這是一難點,二是把不等式問題轉(zhuǎn)化為求函數(shù)的最值,三是在求最值過程中,需求函數(shù)的單調(diào)性,在求單調(diào)性的過程中,要分類討論,這又是一難點,四把問題最后再轉(zhuǎn)化為求不等式.