Discussion

Explanation:

A function is 

Condition 1: Even when, f(x) = f(-x)

Condition 2: Odd when, f(x) = -f(-x)

Given, f(x + y) + f(x - y) = 2f(x)f(y)

Put x = y = 0, we get

f(0) + f(0) = 2f(0)f(0) = 2(f(0))2

⇒ 2f(0) = 2(f(0))2

⇒ f(0) = 1.

Now substituting x = 0, we get

f(y) + f(-y) = 2f(0)f(y)

⇒ f(-y) = f(y)

∴ f(x) satisfies the 1st condition, ; hence it is an even function.

Hence, option (a).

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