Question

已知函数f(x)=1/根号下ax^2-ax+1,(1)若函数f(x)的定义域是R,求实数a的取值范围

已知函数f(x)=1/根号下ax^2-ax+1,(1)若函数f(x)的定义域是R,求实数a的取值范围 (2)若a=2,求函数f(x)的值域

Answer 1

(1) f(x)=1/√(ax²-ax+1)
ax²-ax+1>0
定义域是R
a>0且Δ=a²-4a<0
a(a-4)<0
a-4<0
a<4
a的取值范围：0<a<4
(2) f(x)=1/√(2x²-2x+1)
=1/√(2(x-1/2)²+1/2)
设g(x)=√(2(x-1/2)²+1/2)，当x=1/2时，g(x)min=√(1/2)
f(x)max=1/√(1/2)=√2
值域：(0,√2]

Answer 2

(1) a=0 1>0 ; a>0 a^2-4a<0 0<a<4 0<=a<4
(2)2x^2-2x+1=2(x^2-x)+1=2[(x-1/2)^2-1/4]+1=2(x-1/2)^2-1/2+1=2(x-1/2)^2+1/2>=1/2
1/根号下ax^2-ax+1=1/根号下2x^2-2x+1<=1/根号下1/2=根号2 函数f(x)的值域为

（0，根号2]