【答案】
分析:(1)因為對任意x∈R都有f(x)≤1,所以把函數(shù)變?yōu)轫旤c形式,且a>0,b>0,有當(dāng)x=
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時,f(
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)≤1,化簡即可得證;(2)①先證明必要性:討論絕對值不等式|f(x)|≤1的解集為f(x)≤1或f(x)≥-1,分別得到a的范圍,求出公共解集即可;②證明充分性;由b-1≤a得f(x)≥-1得到f(x)的取值范圍,由a≤2
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.f(x)≤1,求出公共解集得到f(x)的范圍即可.
(3)先證必要性:f(x)≤1得到a-b≤1即a≤b+1;再證充分性:由a≤b+1得到f(x)≤1,得到|f(x)|≤1的充要條件.
解答:(1)證明:根據(jù)題設(shè),對任意x∈R,都有f(x)≤1.
又f(x)=-b(x-
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)
2+
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.∴f(
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)=
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≤1,
∵a>0,b>0,
∴a≤2
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.
(2)證明:必要性:對任意x∈[0,1],|f(x)|≤1⇒f(x)≥-1.據(jù)此可推出f(1)≥-1,即a-b≥-1,∴a≥b-1.
對任意x∈[0,1],|f(x)|≤1⇒f(x)≤1,因為b>1,可得0<
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<1,可推出f(
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)≤1,即a•
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-1≤1,∴a≤2
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,∴b-1≤a≤2
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.
充分性:因為b>1,a≥b-1,對任意x∈[0,1],可以推出ax-bx
2≥b(x-x
2)-x≥-x≥-1,即ax-bx
2≥-1,因為b>1,a≤2
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對任意x∈[0,1],可以推出:ax-bx
2≤2
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x-bx
2-b(x-
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)
2+1≤1,即ax-bx
2≤1,∴-1≤f(x)≤1.
綜上,當(dāng)b>1時,對任意x∈[0,1],|f(x)|≤1的充要條件是b-1≤a≤2
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.
(3)解:因為a>0,0<b≤1時,對任意x∈[0,1]有f(x)=ax-bx
2≥-b≥-1,即f(x)≥-1;
f(x)≤1⇒f(1)≤1⇒a-b≤1,即a≤b+1,
又a≤b+1⇒f(x)≤(b+1)x-bx
2≤1,即f(x)≤1.
所以,當(dāng)a>0,0<b≤1時,對任意x∈[0,1],|f(x)|≤1的充要條件是a≤b+1.
點評:讓學(xué)生理解函數(shù)恒成立時滿足的條件,會找一個命題的充分必要條件.