1>0,f(x)为增函数,),(12422aaax????时,x2-(2+a)x+1<0,f(x)为减函数,),(24212aaax????时,x2-(2+a)x+1<0,f(x)为减函数,),(??????2422aaax时,x2-(2+a)x+1>0,f(x)为增函数;当a<-4时,△>0,方程x2-(2+a)x+1=0有两根2422aaax????,并且024224222????????aaaaaa,∴x?(0,1)时,x2-(2+a)x+1>0,f(x)为增函数,x?(1,+?)时,x2-(2+a)x+1>0,f(x)为增函数。综上所述,当a≤0时,f(x)的递增区间为(0,1),(1,+?);当a>0时,f(x)的递增区间为),(24202aaa???,),(?????2422aaa,递减区间为),(12422aaa???,),(24212aaa???。(2)∵a≥21,∴f(x)在),(24202aaa???上递增,在),(12422aaa???上递减,在),(24212aaa???上递减,在),(?????2422aaa上递增。因此f(x)在(0,1)上有极大值)(2422aaaf???,在(1,+?)上有极小值)(2422aaaf???。即f(b)≤)(2422aaaf???,f(c)≥)(2422aaaf???。因此,f(c)-f(b)≥)(2422aaaf???-)(2422aaaf???=aaaaaaaaaaaaaa424242422222????????????ln=aaaaa4242222?????ln由于a≥21,∴2342??aa,424222lnln????aaa,故f(c)-f(b)>23。评注:这里该不会是根据a≥0得01224222?????lnlnaaa,从而得到f(c)-f(b)≥23的吧。(许毓华2016年10月17日于新宁一中)
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