题目

已知函数f(x)=的定义域为R,则实数a的取值范围为(    )A.a＞             B.-12＜a＜0            C.-12＜a≤0         D.a≤ 答案：C解析:a=0时,定义域为R,排除A、B,当a≠0时,设y=ax2+ax-3,则y值必为R且y≠0,∴Δ=a2+4×3×a＜0.∴-12＜a＜0.综合得-12＜a≤0. Let us not waste____________ time we have left. A. the little       B. little    C. a little    D. a little more