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已知f（x）是定义在区间[﹣1，1]上的奇函数，且f（﹣1）=1，若m，n∈[﹣1，1]，m+n≠0时，有
f
(
m
)
+
f
(
n
)
m
+
n
 ＜0．

（Ⅰ）证明：f（x）在区间[﹣1，1]上是单调减函数；

（Ⅱ）解不等式f（x+

1
2
）＜f（ −
1
x
1
 ）；

（Ⅲ）若f（x）≤t2﹣mt﹣1对所有x∈[﹣1，1]，m∈[0，1]恒成立，求实数t的取值范围．

上一题 下一题 0.0难度 选择题 更新时间:2017-07-07 12:36:49

答案(点此获取答案解析）

证明：（I）令m=x1，n=﹣x2，且﹣1≤x1＜x2≤1，

代入 f(m)+f(n<

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