Question

设函数y=1/根号下ax^2+ax+1的定义域为R,求实数a的取值范围

Answer 1

1,当a=0时,分母为1,函数变为y=1,定义域自然为R

2,当a≠0时,根号下的要恒大于0,即ax^2+ax+1＞0,配方a(x^2+x+0.25)-0.25a+1＞0,即a(x+0.5)^2-0.25a+1＞0 恒大于0则需要a＞0且-0.25a+1＞0即4＞a＞0

综合可知4＞a≥0时,其定义域为R