Let f(x) be a real valued function with domain R such that
f(x + p) = 1 + [2 –3f(x) + 3(f(x)) 2 –(f(x)) 3 ] 1/3 holds good ∀ x ∈ R and some positive constant P, then prove that f(x) is a periodic function.
Text Solution
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Sol. f(x + p) = 1 + [2 –3f(x) + 3(f(x)) 2 –(f(x)) 3 ] 1/3
⇒ f(x + p) = 1 + [1 + (1 – f(x)) 3 ] 1/3 ⇒ f(x + p)–1 = [1 –(f(x) –1)
3 ] 1/3 ⇒ g(x + p) = [1 –(g (x)
3 ] 1/3 … (i)
(g(x) = f(x) –1 (say))
replacing x by x + p
g(x + 2p) = [1 –{g (x + p)} 3 ] 1/3 ⇒ g (x + 2p) = [1 –(1 –g(x))
3 ] 1/3 by (i)
⇒ g(x + 2p) = g(x)
⇒ f(x + 2p) –1 = f(x) –1
⇒ f(x + 2p) = f(x)
Hence f(x) is a periodic function with period 2p.
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