Question

15．已知函數(shù)f（x）=$\sqrt{{x^2}-ax+3a}$，對(duì)于任意x≥2，當(dāng)△x＞0時(shí)，恒有f（x+△x）＞f（x），則實(shí)數(shù)a的取值范圍是[-4，4]．

Answer 1

分析 由題意可得函數(shù)單調(diào)遞增，結(jié)合圖象可得$\left\{\begin{array}{l}{x=-\frac{-a}{2×1}≤2}\\{{2}^{2}-2a+3a≥0}\end{array}\right.$，解關(guān)于a的不等式組可得．

解答 解：∵函數(shù)f（x）=$\sqrt{{x^2}-ax+3a}$，對(duì)于任意x≥2，當(dāng)△x＞0時(shí)，恒有f（x+△x）＞f（x），
∴函數(shù)f（x）=$\sqrt{{x^2}-ax+3a}$在x≥2時(shí)單調(diào)遞增，
∴$\left\{\begin{array}{l}{x=-\frac{-a}{2×1}≤2}\\{{2}^{2}-2a+3a≥0}\end{array}\right.$，解得-4≤a≤4
故答案為：[-4，4]

點(diǎn)評(píng) 本題考查函數(shù)的單調(diào)性和恒成立問(wèn)題，轉(zhuǎn)化為二次函數(shù)的單調(diào)性是解決問(wèn)題的關(guān)鍵，屬中檔題．