An even polynomial function `f(x)` satisfies a relation `f(2x)(1-f(1/(2x)))+f(16x^2y)=f(-2)-f(4x y) \ AAx ,y in R-{0}` and `f(4)=-255 ,f(0)=1.` Then the value of `|(f(2)+1)//2|` is_________

Answer

-15

Explanation

We have `f(2x)-f(2x) f((1)/(2x))+f(16x^(2)y)=f(-2)-f(4xy)`
Replacing y by `(1)/(8x^(2)),` we get
`f(2x)-f(2x)((1)/(2x))+f(2)=f(-2)-f((1)/(2x))`
` :. f(2x)+f((1)/(2x))=f(2x)f((1)/(2x)) ` [As `f(x)` is even]
` :. f(2x)=1+-(2x)^(n)`
`or f(x)=1+-x^(n)`
Now, `f(4)=1+-4^(n)= -255 " "` (Given)
Taking negative sign, we get `256=4^(n) or n=4.`
Hence, `f(x)=1-x^(4)`, which is an even function.
Therefore, `f(2)= -15.`