If-a-function-f-R-R-satisfies-the-relation-f-x-1-f-x-1-3-f-x-for-all-x-R-then-a-period-of-f-is-A-10-B-12-C-6-D-4-

Question Number 119635 by Ar Brandon last updated on 25/Oct/20

If a function f : R \to R satisfies the relation

f (x + 1) + f (x - 1) = \sqrt{3} f (x) for all x \in R

then a period of f is

(A) 10 (B) 12 (C) 6 (D) 4

Answered by Olaf last updated on 26/Oct/20

f (x + 1) + f (x - 1) = \sqrt{3} f (x)

f (x + 2) + f (x) = \sqrt{3} f (x + 1)

f (x + 2) + f (x) = \sqrt{3} [\sqrt{3} f (x) - f (x - 1)]

f (x + 2) = 2 f (x) - \sqrt{3} f (x - 1)

f (x + 3) = 2 f (x + 1) - \sqrt{3} f (x)

f (x + 3) = 2 [\sqrt{3} f (x) - f (x - 1)] - \sqrt{3} f (x)

f (x + 3) = \sqrt{3} f (x) - 2 f (x - 1)

f (x + 4) = \sqrt{3} f (x + 1) - 2 f (x)

f (x + 4) = \sqrt{3} [\sqrt{3} f (x) - f (x - 1)] - 2 f (x)

f (x + 4) = f (x) - \sqrt{3} f (x - 1)

f (x + 5) = f (x + 1) - \sqrt{3} f (x)

f (x + 5) = f (x + 1) - [f (x + 1) + f (x - 1)]

f (x + 5) = - f (x - 1)

f (x + 6) = - f (x)

\Rightarrow f (x + 12) = - f (x + 6)

f (x + 12) = - [- f (x)]

f (x + 12) = f (x)

Period is 12 .

Commented by Ar Brandon last updated on 26/Oct/20

Thank you Sir