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【简答题】
设M是由满足下列条件的函数f（x）构成的集合：“①方程f（x）-x=0有实数根；②函数f（x）的导数f′（x）满足0＜f′（x）＜1”． （Ⅰ）判断函数 f(x)= x 2 + sinx 4 是否是集合M中的元素，并说明理由； （Ⅱ）集合M中的元素f（x）具有下面的性质：若f（x）的定义域为D，则对于任意[m，n]?D，都存在x 0 ∈[m，n]，使得等式f（n）-f（m）=（n-m）f'（x 0 ）成立”，试用这一性质证明：方程f（x）-x=0只有一个实数根； （Ⅲ）设x 1 是方程f（x）-x=0的实数根，求证：对于f（x）定义域中任意的x 2 、x 3 ，当|x 2 -x 1 |＜1，且|x 3 -x 1 |＜1时，|f（x 3 ）-f（x 2 ）|＜2．
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