问题 解答题
设函数f(x)=ax-(a+1)ln(x+1),其中a>0.
(Ⅰ)求f(x)的单调区间;
(Ⅱ)当x>0时,证明不等式:
x
1+x
<ln(x+1)<x

(Ⅲ)设f(x)的最小值为g(a),证明不等式:-
1
a
<g(a)<0
答案

(Ⅰ)∵f(x)=ax-(a+1)ln(x+1),其中a>0,

∴函数f(x)的定义域为(-1,+∞),且f(x)=

ax-1
x+1
,a>0,

由f′(x)=0,得x=

1
a

当x变化时,f′(x),f(x)的变化情况如下表:

 x (-1,
1
a
 
1
a
 (
1
a
,+∞)
 f′(x)- 0+
 f(x) 极小值
由上表知,当x∈(-1,
1
a
)时,f′(x)<0,函数f(x)在(-1,
1
a
)内单调递减;

当x∈(

1
a
,+∞)时,f′(x)>0,函数f(x)在(
1
a
,+∞
)内单调递增.

∴函数f(x)的增区间是(

1
a
,+∞),减区间是(-1,
1
a
).

(Ⅱ)证明:设∅(x)=ln(x+1)-

x
1+x
,x∈[0,+∞),

对∅(x)求导,得∅′(x)=

1
x+1
-
1
(1+x)2
=
x
(1+x)2

当x≥0时,∅′(x)≥0,所以∅(x)在[0,+∞)内是增函数.

∴∅(x)>∅(0)=0,即ln(x+1)-

x
1+x
>0,

x
1+x
<ln(x+1).

同理可证ln(x+1)<x,

x
1+x
<ln(x+1)<x.

(Ⅲ)由(Ⅰ)知,g(a)=f(

1
a
)=1-(a+1)•ln(
1
a
+1),

x=

1
a
代入
x
1+x
<ln(x+1)<x

1
a+1
<ln(
1
a
+1)<
1
a

即1<(a+1)ln(

1
a
+1)<1+
1
a

-

1
a
<1-(a+1)ln(
1
a
+1)<0,

故-

1
a
<g(a)<0.

选择题
阅读理解

阅读理解。

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