function

ans is a,b,c,d .

put x²=z

f(z) + f(√z) =2

f(x) + f(√x) =2

thus f(x²) = f(x^1/2)

similarly doing this recursively,

f(x²) = f(x^1/n) (n→∞) = f(1) = constant

how can u call a constant function as periodic??
if so then wat is itz period..???
It has an undefined period(infinitesmial)..

Aditya, you can make your fine solution a little more rigorous.

First note that

f(0) = f(1)=1 can be directly obtained. By the way this means the function is not injective.

Further

f(x)+f(x^2) = f(-x)+f(x^2) \Rightarrow f(x) = f(-x)

Hence it suffices to investigate the function for x≥0.

To proceed as Aditya has done, we need an additional assumption that f is continuous (or maybe it can be proved from the equation somehow, but the point is you need f to be continuous to take those limits inside the bracket).

Suppose 0<x<1

f(x) = f(x^4)=...= f(x^{4n})=..

implies that f(x) = \lim_{n \rightarrow \infty} f(x^{4n})= f \left(\lim_{n \rightarrow \infty} x^{4n} \right) = f(0) =1

Suppose x>1, then

f(x) = f(x^{\frac{1}{4}})=...= f(x^{\frac{1}{4n}})=..

implies that

f(x) = \lim_{n \rightarrow \infty} f(x^{\frac{1}{4n}})= f \left(\lim_{n \rightarrow \infty} x^{\frac{1}{4n}} \right) = f(1) =1

@hsbhatt sir - thanks sir.

@ayush - a constant function is periodic though its period is undefined.