Letf(x,y) be a periodic function satisfying the conditionf(x,y) =f((2x+ 2y), (2y− 2x)) ∀x,y∈R.Now define a functiongbyg(

Solution

f (x, y) = f (2x + 2y, 2y − 2x) …(i)

⇒ f (2x + 2y, 2y − 2x) = f {2(2x + 2y) + 2(2y − 2x), 2(2y − 2x) − 2(2x + 2y)}

⇒ f (x, y) = f (8y − 8x)    [using (i)] …(ii)

⇒ f (8y, −8x) = f {8(−8x), −8(8y)}        [using (ii)]

⇒ f (x, y) = f (2x + 2y, 2y − 2x) = f (8y, −8x) = f (−64x, −64y)

⇒ f (x, y) = f (−64x, −64y) …(iii)

⇒ f (−64x, −64y) = f (64 × 64x, 64 × 64y) = f (212x, 212y)

⇒ f (x, y) = f (212x, 212y)        [using (iii)]

⇒ f (2x, 0) = f (212 . 2x, 0) = f (212+x , 0) …(iv)

given g(x, 0) = f (2x, 0)  ⇒ g(x, 0) = f (2x, 0) = f (212, 0)         [using (iv)]

⇒ g(x, 0) = g(x + 12, 0).  Hence, g(x) is periodic with period 12.