已知函数f(x)=x
2-(a+2)x+alnx.其中常数a>0.
(1)当a>2时,求函数f(x)的单调递增区间;
(2)当a=4时,给出两类直线:6x+y+m=0与3x-y+n=0,其中m,n为常数,判断这两类直线中是否存在y=f(x)的切线,若存在,求出相应的m或n的值,若不存在,说明理由.
(3)设定义在D上的函数y=h(x)在点P(x
,h(x
))处的切线方程为l:y=g(x),当x≠x
时,若
![manfen5.com 满分网](http://img.manfen5.com/res/GZSX/web/STSource/20131103100003398559872/SYS201311031000033985598021_ST/0.png)
在D内恒成立,则称P为函数y=h(x)的“类对称点”,当a=4时,试问y=f(x)是否存在“类对称点”,若存在,请至少求出一个“类对称点”的横坐标,若不存在,说明理由.
考点分析:
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