It is known that the function f (x) whose domain is r has f (x + y) + F (X-Y) = 2F (x) f (y) for any real number x and y And f (x) ≠ 0 prove that f (x) is even

It is known that the function f (x) whose domain is r has f (x + y) + F (X-Y) = 2F (x) f (y) for any real number x and y And f (x) ≠ 0 prove that f (x) is even

Substituting x = 0, y = 0 into f (0 + 0) + F (0-0) = 2F (0) f (0)
That is, 2f (0) = 2F (0) &# 178;, ∵ f (0) ≠ 0, ∵ f (0) = 1
Substituting x = 0 into f (0 + y) f (0-y) = 2F (0) f (y)
That is, f (y) + F (- y) = 2F (y) ‖ f (y) = f (- y)
F (x) is an even function

Let f (x) be defined as R. for any real number α, β, f (α) + F (β) = 2F ((α + β) / 2) * f ((α - β) / 2), f (π / 3) = 1 / 2, f (π / 2) = 0
1. F (0), ((2 π) / 2)
2. Proof: F (- x) = f (x) = - f (π - x)
3. If 0 ＜ = x ＜ = π / 2, f (x) > 0, it is proved that f (x) decreases monotonically on [0, π]
4. Find the minimum positive period of F (x)
Urgent!!!!!!!!!!

1, let α = β = π / 3, substitute to get 2F (π / 3) = 2F (π / 3) × f (0), get f (0) = 1, then let α = π, β = 0, have f (0) + F (π) = 2F (π / 2) × f (π / 2), get f (π) = 12, let α = x, β = - x, substitute to have f (x) + F (- x) = 2F (x) × f (0), f (0) = 1, get f (- x) = f (x) × f (0)

Let f (x) be a function defined on (1, + ∞), and f (x) = 2F (1 / x) √ X-1, find f (x)
It is said in the book that 1 / X can replace X. This is understandable, but I think there is something wrong with the domain of definition
The domain of F (x) is (1, + ∞), so the domain of F (x) = 2F (1 / x) √ X-1 should be x > 1 and 1 / x > 1, and there is no solution to this inequality system. Therefore, the problem is wrong
Excuse me this topic is really wrong, I Baidu, many people do not see, directly into the generation?

Your query is correct. If we only consider the domain of definition, the topic will be problematic