Question

Let a real valued function f(x) satisfying f(x + y) + f(x – y) = f(x)f(y) {f(0) ≠ 0} ∀ x, y ∈ R, then f(–2) – f(–1) + f(0) + f(1) – f(2) is equal to ______.

पर्याय

• 0.00

• 1.00

• 2.00

• 3.00

Answer 1

Let a real valued function f(x) satisfying f(x + y) + f(x – y) = f(x)f(y) {f(0) ≠ 0} ∀ x, y ∈ R, then f(–2) – f(–1) + f(0) + f(1) – f(2) is equal to 2.00.

Explanation:

Put x = y = 0

2f(0) = f²(0)

⇒ f(0) = 2

Now put x = 0

f(–y) + f(y) = f(0)f(y)

⇒ f(y) = f(–y)

⇒ f(–2) – f(–1) + f(0) + f(1) – f(2) = f(0) = 2