f(x)=ax²+x-1+3a(a属于R)在区间 [-1,1]上有零点
即ax²+x-1+3a=0在[-1,1]上有实数解
即a(x²+3)=1-x
即 a=(1-x)/(x²+3)有实数解
令 g(x)=(1-x)/(x²+3)则
a的范围即是g(x)的值域
g'(x)=[-x²-3-2x(1-x)]/(x²+3)²
=(x²-2x-3)/(x²+3)²
=(x+1)(x-3)/(x²+3)²
∵-1≤x≤1∴ (x+1)(x-3) ≤0
∴ g'(x)≤0
∴g(x)是减函数
∴x=-1 g(x)max=1/2
x=1,g(x)min=0
∴g(x)值域为[0,1/2]
∴实数a的取值范围是[0,1/2]