function - a das gupta

See ... 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 ...

jangra 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

this question is correct. u can check in a das gupta book and also in fiitjee packages.

Qwerty ... 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 ...

see 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