证:(1) ∵ tÎR, t ¹ –1,
∴ ⊿ = (–c
2a)
2 – 16c
2 = c
4a
2 – 16c
2³ 0 ,
∵ c ¹ 0, ∴c
2a
2³ 16 , ∴| ac | ³ 4.
(2) 由 f ( x ) =" 1" –
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823144858170271.gif)
,
法1. 设–1 < x
1 < x
2, 则f (x
2) – f ( x
1) =" 1–"
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823144858248293.gif)
–1 +
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823144858264280.gif)
=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823144858279620.gif)
.
∵ –1 < x
1 < x
2, ∴ x
1 – x
2 < 0, x
1 + 1 > 0, x
2 + 1 > 0 ,
∴f (x
2) – f ( x
1) < 0 , 即f (x
2) < f ( x
1) , ∴x ³ 0时,f ( x )单调递增.
法2. 由f ` ( x ) =
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823144858295436.gif)
> 0 得x ¹ –1, ∴x > –1时,f ( x )单调递增.
(3)∵f ( x )在x > –1时单调递增,| c | ³
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823144858357242.gif)
> 0 ,
∴f (| c | ) ³ f (
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823144858357242.gif)
) =
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823144858389380.gif)
=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823144858404291.gif)
f ( | a | ) + f ( | c | ) =
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823144858435302.gif)
+
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823144858404291.gif)
>
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823144858482309.gif)
+
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823144858404291.gif)
=1.
即f ( | a | ) + f ( | c | ) > 1.