1this question is correct. u can check in a das gupta book and also in fiitjee packages.
If f(x)\epsilon[1,2] when x \epsilon R and for a fixedreal number p
f(x+p) = 1 + √f(x)- {f(x)}2 for all x \epsilon R then prove that f(x) is a peroidic function.
-
UP 0 DOWN 0 0 6
6 Answers
1See ... I'm not sure of my solution ... still I'm trying ...
For func. to be periodic .... f(x) = f(x+p)
Therefore ... f(x+p)ε[1,2]
0 <= √[f(x) - {f(x)}2 ] <= 1
0 <= f(x) - {f(x)}2 <= 1
Now .. u can try ...
23jangra are u sure question is correct ??
f - f2 ≥ 0
0 ≤ f ≤ 1 but it is given that f ≡ [1,2] ???
so only 1 value of f is possible that is f = 1 ???
btw arka it isnt mentioned that period is p
11Qwerty ... ur right ... that's why I said I wasn't sure of the solution ...
But it seems that [ seeing the question pattern ] the func. is periodic over p ...
23see jangra there r 2 things ,
1] if u want the solution to the question u typed , then only possible thing is f = 1
so f is a periodic function
2] if u dont agree to this , then u will hav to agree that u hav typed the question wrong
it shud be f(x+p) = 1/2 + √f(x) -f(x)2
so(f(x+p)-1/2)^{2} =f(x)-f(x)^{2}
f(x+p)^{2}-f(x+p) =f(x)-f(x)^{2}- 1/4
f(x+p)-f(x+p)^{2} =(f(x)-1/2) ^{2}........(1)
then put x = x+p in the functional eqn given
f(x+2p)=1/2 + \sqrt{f(x+p)[1-f(x+p)]}
f(x+2p)=1/2 +\sqrt{f(x+p)-f(x+p)^{2}}=1/2+f(x)-1/2=f(x).from 1
hence period = 2p
1QWERTY is right ... either there's a typing mistake ... or the func. is periodic over 2p ...