If the image of the function f (x) = (x-2a) (x + A ^ 2) is symmetric about the Y axis, then a =?
On y symmetry, that is even function, then f (x) = f (- x)
(x-2a)(x+a^2)=(-x-2a)(-x+a^2)
x^2+(a^2-2a)x-2a^3=x^2+(2a-a^2)x-2a^3
a^2-2a=2a-a^2
a(a-2)=0
So a = 0, 2
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