(1)解:f"(x)=lnx+1(x>0),令f"(x)=0,得.(2分)
∵当时,f"(x)<0;当时,f"(x)>0,
∴当时,.
(2)F(x)=ax2+lnx+1(x>0),.
①当a≥0时,恒有F"(x)>0,F(x)在(0,+∞)上是增函数;
②当a<0时,令F"(x)>0,得2ax2+1>0,解得;
令F"(x)<0,得2ax2+1<0,解得.
综上,当a≥0时,F(x)在(0,+∞)上是增函数;当a<0时,F(x)在上单调递增,在上单调递减.
(3)证:.
要证,即证,
等价于证,
令,
则只要证,由t>1知lnt>0,故等价于证lnt<t﹣1<tlnt(t>1)(*).
①设g(t)=t﹣1﹣lnt(t≥1),则,
故g(t)在[1,+∞)上是增函数,
∴当t>1时,g(t)=t﹣1﹣lnt>g(1)=0,即t﹣1>lnt(t>1).
②设h(t)=tlnt﹣(t﹣1)(t≥1),则h"(t)=lnt≥0(t≥1),
故h(t)在[1,+∞)上是增函数,
∴当t>1时,h(t)=tlnt﹣(t﹣1)>h(1)=0,即t﹣1<tlnt(t>1).
由①②知(*)成立,得证.
© 2017-2019 超级试练试题库,All Rights Reserved.