Question

16．设函数$f（x）=\left\{\begin{array}{l}\int_1^e{\frac{1}{t}dt，x＞\sqrt{2}}\\ \frac{1}{3}，x≤\sqrt{2}\end{array}\right.$，若$f（{x_0}）＞\frac{1}{2}$，则x₀的取值范围为x₀＞$\sqrt{2}$．

Answer 1

分析 x＞$\sqrt{2}$，f（x）=lnx|${\;}_{1}^{e}$=1，利用$f（{x_0}）＞\frac{1}{2}$，可得x₀的取值范围．

解答 解：x＞$\sqrt{2}$，f（x）=lnx|${\;}_{1}^{e}$=1，
∵$f（x）=\left\{\begin{array}{l}\int_1^e{\frac{1}{t}dt，x＞\sqrt{2}}\\ \frac{1}{3}，x≤\sqrt{2}\end{array}\right.$，$f（{x_0}）＞\frac{1}{2}$，
∴x₀＞$\sqrt{2}$，
故答案为x₀＞$\sqrt{2}$．

点评 本题考查分段函数，考查不等式的解法，考查学生的计算能力，属于中档题．