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已知x∈R，函数f（x）=ax 3 +bx 2 +cx+d在x=0处取得极值，曲线y=f（x）过原点O（0，0）和点P（-1，2）．若曲线y=f（x）在点P处的切线l与直线y=2x的夹角为45°，且直线l的倾斜角θ∈（ π 2 ，π）， （Ⅰ）求f（x）的解析式； （Ⅱ）若函数y=f（x）在区间[2m-1，m+1]上是增函数，求实数m的取值范围； （Ⅲ）若x1、x 2 ∈[-1，1]，求证：f（x 1 ）-f（x 2 ）≤4．
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