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2024届邯郸市高三第四次调研考试数学试卷答案

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分析利用函数奇偶性和单调性之间的关系得到不等式f（x）＞0和f（x）＜0的解，然后将不等式（x-2）•f（x）＜0转化为$\left\{\begin{array}{l}{x-2＞0}\\{f（x）＜0}\end{array}\right.$①或$\left\{\begin{array}{l}{x-2＜0}\\{f（x）＞0}\end{array}\right.$，②，进行求解．

解答解：∵f（x）是偶函数，且在[0，+∞）内是增函数，
∴f（x）在（-∞，0]内是减函数，
∵f（-3）=-f（3）=0，
∴f（3）=0．
则f（x）对应的图象如图：
则不等式（x-2）•f（x）＜0等价为：
$\left\{\begin{array}{l}{x-2＞0}\\{f（x）＜0}\end{array}\right.$①或$\left\{\begin{array}{l}{x-2＜0}\\{f（x）＞0}\end{array}\right.$，②
由①得$\left\{\begin{array}{l}{x＞2}\\{-3＜x＜3}\end{array}\right.$，得2＜x＜3．
由②得$\left\{\begin{array}{l}{x＜2}\\{x＞3或x＜-3}\end{array}\right.$，得x＜-3．
综上：2＜x＜3或x＜-3．
故不等式的解集为：（-∞，-3）∪（2，3），
故选：A

点评本题主要考查函数奇偶性和单调性之间的关系的应用，利用数形结合是解决本题的关键．