Question

16．設函數$f（x）=\left\{\begin{array}{l}\int_1^e{\frac{1}{t}dt，x＞\sqrt{2}}\\ \frac{1}{3}，x≤\sqrt{2}\end{array}\right.$，若$f（{x_0}）＞\frac{1}{2}$，則x₀的取值范圍為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的取值范圍．

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＞$\sqrt{2}$，
故答案為x₀＞$\sqrt{2}$．

點評 本題考查分段函數，考查不等式的解法，考查學生的計算能力，屬于中檔題．