Question

20．已知函數(shù)f（x）=$\left\{\begin{array}{l}{-{x}^{2}-ax-5（x≤1）}\\{\frac{a}{x}（x＞1）}\end{array}\right.$滿足對(duì)任意x₁≠x₂，都有$\frac{f（{x}_{1}）-f（{x}_{2}）}{{x}_{1}-{x}_{2}}$＞0成立，則a的范圍是（　　）

• A． -3≤a＜0
• B． -3≤a≤-2
• C． a≤-2
• D． a≤0

Answer 1

分析 由已知可得：函數(shù)f（x）=$\left\{\begin{array}{l}{-{x}^{2}-ax-5（x≤1）}\\{\frac{a}{x}（x＞1）}\end{array}\right.$在R上為增函數(shù)，進(jìn)而$\left\{\begin{array}{l}{-\frac{a}{2}≥1}\\{a＜0}\\{a≥-1-a-5}\end{array}\right.$，解得a的取值范圍．

解答 解：對(duì)任意的x₁≠x₂，都有$\frac{f（{x}_{1}）-f（{x}_{2}）}{{x}_{1}-{x}_{2}}$＞0成立，
則函數(shù)f（x）=$\left\{\begin{array}{l}{-{x}^{2}-ax-5（x≤1）}\\{\frac{a}{x}（x＞1）}\end{array}\right.$在R上為增函數(shù)，
∴$\left\{\begin{array}{l}{-\frac{a}{2}≥1}\\{a＜0}\\{a≥-1-a-5}\end{array}\right.$，∴-3≤a≤-2．
故選B．

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．