解:(1)
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①當2<a<8時,當
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時,f'(x)<0,∴f(x)在
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單調遞減;
當
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時,f'(x)>0,∴f(x)在
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單調遞增;
②當a≤2時,f'(x)≥0,∴f(x)在(1,2)單調遞增;
③當a≥8時,f'(x)≤0,∴f(x)在(1,2)單調遞減;
(2)
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,依題意f'(x)≥0,x∈(1,2],即a≤2x
2,x∈(1,2].
∵上式恒成立,∴a≤2.①
又
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,依題意g'(x)≤0,x∈(0,1),即
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,x∈(0,1).
∵上式恒成立,∴a≥2.②
由①②得a=2
∴方程f(x)=g(x)+2,
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.
設
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,
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令h'(x)>0,并由x>0,得
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,解知x>1
令h'(x)<0,由x>0,解得0<x<1
列表分析:
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知h(x)在x=1處有一個最小值0
當x>0且x≠1時,h(x)>0,
∴h(x)=0在(0,+∝)上只有一個解.
即當x>0時,方程f(x)=g(x)+2有唯一解;
(3)f(x)
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在(0,1]上恒成立,
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在(0,1]上恒成立.
設
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,則
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,
∵0<x≤1?x
2-2<0,2lnx<0,
∴H′(x)<0,H(x)d (0,1]單調遞減,
∴-1<b≤1,又∵b>-1,∴
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.
分析:(1)由f(x)的解析式求出f(x)的導函數,分a≤2和a≥8以及2<a<8三種情況,分別令導函數大于0列出關于x的不等式,求出不等式的解集即可得到x的范圍即為函數的遞增區(qū)間;令導函數小于0列出關于x的不等式,求出不等式的解集即可得到x的范圍即為函數的遞減區(qū)間;
(2)由(1)不難給出方程f(x)=g(x)+2,然后構造函數,利用函數的單調性證明方程解的唯一性;
(3)f(x)
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在(0,1]上恒成立
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在(0,1]上恒成立.由此能導出b的取值范圍.
點評:利用導數研究函數的單調性比用函數單調性的定義要方便,但應注意f′(x)>0(或f′(x)<0)僅是f(x)在某個區(qū)間上為增函數(或減函數)的充分條件,在(a,b)內可導的函數f(x)在(a,b)上遞增(或遞減)的充要條件應是f′(x)≥0[或f′(x)≤0],x∈(a,b)恒成立,且f′(x)在(a,b)的任意子區(qū)間內都不恒等于0,這就是說,函數f(x)在區(qū)間上的增減性并不排斥在區(qū)間內個別點處有f′(x
0)=0,甚至可以在無窮多個點處f′(x
0)=0,只要這樣的點不能充滿所給區(qū)間的任何一個子區(qū)間,因此,在已知函數f(x)是增函數(或減函數)求參數的取值范圍時,應令f′(x)≥0[或f′(x)≤0]恒成立,解出參數的取值范圍(一般可用不等式恒成立理論求解),然后檢驗參數的取值能否使f′(x)恒等于0,若能恒等于0,則參數的這個值應舍去,若f′(x)不恒為0,則由f′(x)≥0[或f′(x)≤0]恒成立解出的參數的取值范圍確定,屬難題.