If the function f(x) satisfies the relation f(x + y) + f(x - y) = 2f(x)f(y), such that x and y belong to the set of real numbers, and f(0) ≠ 0, then f(x) is
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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