学习资料精品资料
学习资料精品资料 则h'(x )=1+ln(x+1).
令h'(x )=0,得x=-1,
易得h (x )在区间内单调递减,在区间内单调递增.
所以h (x )min =h =1->0,∴f'(x )<0.
故f (x )的单调递减区间为(-1,0),(0,+∞).
(2)当x>0时,f (x )>
恒成立, 则k<(x+1)f (x ).
令g (x )=(x+1)f (x )=,则g'(x )=
令φ(x )=1-x+ln(x+1)(x>0)?φ'(x )=-<0,所以φ(x )在区间(0,+∞)内单调递减. 又φ(2)=ln 3-1>0,φ(3)=2ln 2-2<0,
则存在实数t ∈(2,3),使φ(t )=0?t=1+ln(t+1).
所以g (x )在区间(0,t )内单调递减,在区间(t ,+∞)内单调递增.
所以g (x )min =g (t )
=
=t+1∈(3,4),故k max =3.
14.解 (1)因为f (1)=1-=0,所以a=2.
此时f (x )=ln x-x 2+x ,x>0.
则f'(x )=-2x+1=(x>0).
令f'(x )<0,则2x 2-x-1>0.
又x>0,所以x>1.
所以f (x )的单调递减区间为(1,+∞).
(2)(方法一)令g (x )=f (x )-(ax-1)=ln x-ax 2+(1-a )x+1,则g'(x )=-ax+(1-a )=
当a ≤0时,因为x>0,所以g'(x )>0.
所以g (x )在区间(0,+∞)内是增函数, 又g (1)=ln 1-a×12+(1-a )+1=-a+2>0,所以关于x 的不等式f (x )≤ax-1不能恒成立. 当a>0时,g'(x )==-(x>0), 令g'(x )=0,得x=
所以当x 时,g'(x )>0;当x 时,g'(x )<0,
因此函数g (x )在x 内是增函数,在x 内是减函数.
故函数g (x )的最大值为g =ln a
+(1-a )+1=-ln a.