Question

Given f an odd function periodic with period 2 continuous "

• A. g(x) is odd function
• B. g(2n) = 1
• C. g(2n) = 0
• D. None of these

Answer 1

Answer: (C)

g(2n) = 0

Answer 2

g(x + 2) = $\int_{0}^{x + 2}{f(t)}$ dt

$\int_{0}^{2}{f(t)}$dt +$\int_{2}^{x + 2}{f(t)}$dt = g(2) + $\int_{0}^{x}{f(t)}$dt

\ g(x + 2) = g(2) + g(x) Ž g(x) is periodic with period 2

Also, g(2) =$\int_{0}^{2}{f(t)}$dt = $\int_{0}^{1}{f(t)}$dt + $\int_{1}^{2}{f(t)}$dt

= $\int_{0}^{1}{f(t)}$dt + $\int_{–1}^{0}{f(t)}$dt

[putting t = u + 2]

= $\int_{–1}^{1}{f(t)}$dt = 0 [Q f(x) is odd]

\ g(2n) = 0 [Q g(x) is periodic with period 2]