【答案】
分析:(1)先對(duì)函數(shù)f(x)進(jìn)行求導(dǎo),然后令導(dǎo)函數(shù)大于0(或小于0)求出x的范圍,根據(jù)f′(x)>0求得的區(qū)間是單調(diào)增區(qū)間,f′(x)<0求得的區(qū)間是單調(diào)減區(qū)間,即可得到答案.
(2))處的切線(xiàn)的傾斜角為45°,得到f′(2)=1求出a的值代入到
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中化簡(jiǎn),求出導(dǎo)函數(shù),因?yàn)楹瘮?shù)在(2,3)上總存在極值得到
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解出m的范圍記即可;(3)是近年來(lái)高考考查的熱點(diǎn)問(wèn)題,即與函數(shù)結(jié)合證明不等式問(wèn)題,常用的解題思路是利用前面的結(jié)論構(gòu)造函數(shù),利用函數(shù)的單調(diào)性,對(duì)于函數(shù)取單調(diào)區(qū)間上的正整數(shù)自變量n有某些結(jié)論成立,進(jìn)而解答出這類(lèi)不等式問(wèn)題的解.
解答:解:(1)
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,
當(dāng)a>0時(shí),f(x)的單調(diào)增區(qū)間為(0,1],減區(qū)間為[1,+∞);
當(dāng)a<0時(shí),f(x)的單調(diào)增區(qū)間為[1,+∞),減區(qū)間為(0,1];
當(dāng)a=0時(shí),f(x)不是單調(diào)函數(shù)
(2)因?yàn)楹瘮?shù)y=f(x)的圖象在點(diǎn)(2,f(2))處的切線(xiàn)的傾斜角為45°,
所以f′(2)=1,所以a=-2,
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,
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,g′(x)=3x
2+(4+m)x-2
因?yàn)閷?duì)于任意的t∈[1,2],函數(shù)
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在區(qū)間(t,3)上
總存在極值,所以只需
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,解得
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(3)令a=-1(或a=1)
此時(shí)f(x)=-lnx+x-3,
所以f(1)=-2,
由(1)知f(x)=-lnx+x-3,在[1,+∞)上單調(diào)遞增,
∴當(dāng)x∈(1,+∞)時(shí)f(x)>f(1),即-lnx+x-1>0,
∴l(xiāng)nx<x-1對(duì)一切x∈(1,+∞)成立,
∵n≥2,n∈N
*,
則有
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,
∴要證ln(2
2+1)+ln(3
2+1)+ln(4
2+1)+…+ln(n
2+1)<1+2lnn!
即要證
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,
而
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=1-
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<1.
點(diǎn)評(píng):此題是個(gè)難題.本題考查利用函數(shù)的導(dǎo)數(shù)來(lái)求函數(shù)的單調(diào)區(qū)間,已知函數(shù)曲線(xiàn)上一點(diǎn)求曲線(xiàn)的切線(xiàn)方程即對(duì)函數(shù)導(dǎo)數(shù)的幾何意義的考查,考查求導(dǎo)公式的掌握情況.含參數(shù)的數(shù)學(xué)問(wèn)題的處理,構(gòu)造函數(shù)求解證明不等式問(wèn)題.以及考查學(xué)生創(chuàng)造性的分析解決問(wèn)題的能力.