Question

1．已知函數(shù)在R上是單調(diào)函數(shù)，f（x）=$\left\{\begin{array}{l}{（3a-1）x+4a，x＜1}\\{lo{g}_{a}x，x≥1}\end{array}\right.$，則實(shí)數(shù)a的取值范圍是$\frac{1}{7}$≤a＜$\frac{1}{3}$．

Answer 1

分析 根據(jù)分段函數(shù)單調(diào)性的性質(zhì)進(jìn)行求解即可．

解答 解：∵函數(shù)f（x）是在R上是單調(diào)函數(shù)，
∴若函數(shù)為單調(diào)遞增函數(shù)，
則當(dāng)x≥1和x＜0時，分別得到遞增，且滿足3a-1+4a≤log_a1，
即$\left\{\begin{array}{l}{a＞1}\\{3a-1＞0}\\{7a-1≤0}\end{array}\right.$，則$\left\{\begin{array}{l}{a＞1}\\{a＞\frac{1}{3}}\\{a≤\frac{1}{7}}\end{array}\right.$．此時無解．
若函數(shù)為單調(diào)遞減函數(shù)，
則當(dāng)x≥1和x＜0時，分別得到遞減，且滿足3a-1+4a≥log_a1，
即$\left\{\begin{array}{l}{0＜a＜1}\\{3a-1＜0}\\{7a-1≥0}\end{array}\right.$，則$\left\{\begin{array}{l}{0＜a＜1}\\{a＜\frac{1}{3}}\\{a≥\frac{1}{7}}\end{array}\right.$．解得$\frac{1}{7}$≤a＜$\frac{1}{3}$，
綜上$\frac{1}{7}$≤a＜$\frac{1}{3}$，
故答案為：$\frac{1}{7}$≤a＜$\frac{1}{3}$．

點(diǎn)評 本題主要考查函數(shù)單調(diào)性的應(yīng)用，根據(jù)分段函數(shù)單調(diào)性的性質(zhì)是解決本題的關(guān)鍵．