When x & gt=a
f(x)=2x^2+(x-a)^2=3x^2-2ax+a^2
f(0)=a^2>; = 1
a & lt=- 1
When x
f(x)=2x^2-(x-a)^2=x^2+2ax-a^2
f(0)=-a^2>; = 1
No answer
So a < =- 1
When x & gt=a
f(x)=3x^2-2ax+a^2
When a & lt=0
When x takes a/3, the minimum f(x) is 2/3a 2, and a = 0.
When a & gt=0
When x takes a, the minimum f(x) is 2a 2, and a = 0.
f(x)=0
When x
f(x)=x^2+2ax-a^2
When a & gt=0
When x takes -a, the minimum value of f(x) is-2a 2.
When a & lt=0
When x takes a, the minimum f(x) is 2a 2, and a = 0.
Therefore, the minimum value of f(x) is-2a 2, a >;; 0
x & gta,h(x)=3x^2-2ax+a^2>; = 1
If a 2 >; =3/2
The solution set is (a, +∞)
If a 2
The solution set is [(-∞, (a-√3-2a 2)/3)∩((a+√3-2a 2)/3, +∞)] ∩ (a,+∞).