设函数f（x）=|x﹣1|+12|x﹣3|（Ⅰ）求不等式f（x）＞2的解集；（Ⅱ）若不等式f（x）≤a（x+12）的解集非空，求实数a的取值范围．_刷题宝

题干

设函数f（x）=|x﹣1|+

1

2

|x﹣3|

（Ⅰ）求不等式f（x）＞2的解集；

（Ⅱ）若不等式f（x）≤a（x+

1

2

）的解集非空，求实数a的取值范围．

上一题下一题 0.0难度选择题更新时间:2016-09-06 06:20:30

答案(点此获取答案解析）

解：（ I）去绝对值可得f（x）=|x﹣1|+

1

2

|x﹣3|=

\{\begin{matrix} - \frac{3}{2} \end{matrix}

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