Question

4．函數(shù)f（x）=$\left\{\begin{array}{l}{a{x}^{2}+x-1（x＞2）}\\{ax-1（x≤2）}\end{array}\right.$是R上的減函數(shù)，則實(shí)數(shù)a的取值a范圍（　�。�

• A． [-$\frac{1}{2}$，0）
• B． （-∞，$-\frac{1}{4}$]
• C． [-1，-$\frac{1}{4}$]
• D． （-∞，-1]

Answer 1

分析 若函數(shù)f（x）=$\left\{\begin{array}{l}{a{x}^{2}+x-1（x＞2）}\\{ax-1（x≤2）}\end{array}\right.$是R上的減函數(shù)，則$\left\{\begin{array}{l}a＜0\\-\frac{1}{2a}≤2\\ 2a-1≥4a+2-1\end{array}\right.$，解得實(shí)數(shù)a的取值a范圍

解答 解：∵函數(shù)f（x）=$\left\{\begin{array}{l}{a{x}^{2}+x-1（x＞2）}\\{ax-1（x≤2）}\end{array}\right.$是R上的減函數(shù)，
∴$\left\{\begin{array}{l}a＜0\\-\frac{1}{2a}≤2\\ 2a-1≥4a+2-1\end{array}\right.$，
解得：a∈（-∞，-1]，
故選：D

點(diǎn)評(píng) 本題考查的知識(shí)點(diǎn)是分段函數(shù)的應(yīng)用，正確理解分段函數(shù)的單調(diào)性是解答的關(guān)鍵．