Question

17．已知函數(shù)f（x）=$\left\{\begin{array}{l}{lo{g}_{\frac{1}{3}}x，x＞0}\\{{2}^{x}，x≤0}\end{array}\right.$，若f（log₂$\frac{\sqrt{2}}{2}$）+f[f（9）]=$\frac{1+2\sqrt{2}}{4}$；若f（f（a））≤1，則實數(shù)a的取值范圍是${log}_{2}\frac{1}{3}≤a≤（\frac{1}{3}）^{\frac{1}{3}}$，或a≥1．

Answer 1

分析 根據(jù)已知中函數(shù)f（x）=$\left\{\begin{array}{l}{lo{g}_{\frac{1}{3}}x，x＞0}\\{{2}^{x}，x≤0}\end{array}\right.$，代和計算可得答案．

解答 解：∵函數(shù)f（x）=$\left\{\begin{array}{l}{lo{g}_{\frac{1}{3}}x，x＞0}\\{{2}^{x}，x≤0}\end{array}\right.$，
∴f（log₂$\frac{\sqrt{2}}{2}$）+f[f（9）]=f（-$\frac{1}{2}$）+f（-2）=$\frac{1+2\sqrt{2}}{4}$，
若f（f（a））≤1，
則f（a）≤0，或f（a）$≥\frac{1}{3}$，
∴${log}_{2}\frac{1}{3}≤a≤（\frac{1}{3}）^{\frac{1}{3}}$，或a≥1，
故答案為：$\frac{1+2\sqrt{2}}{4}$，${log}_{2}\frac{1}{3}≤a≤（\frac{1}{3}）^{\frac{1}{3}}$，或a≥1．

點評 本題考查的知識點是分段函數(shù)的應(yīng)用，方程思想，對數(shù)的運算性質(zhì)，難度中檔．