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【简答题】
已知函数 f x =ln|x| x≠0 ，函数 g x = 1 f′ x +af′ x x≠0 （I）当x≠0时，求函数 y=g x 的表达式； （Ⅱ）若a＞0，且函数 y=g x 在 0，+∞ 上的最小值是2，求a的值； （Ⅲ）对于（Ⅱ）中所求的a值，若函数 h(x)= 1 3 x 3 - b+1 2a x 2 +bx，x∈R ，恰有三个零点，求b的取值范围．
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