f(x) is a continuous fxn.
f(x)+f(2x+y)+5xy=f(3x-y)+2x^{2}+1 for all x belongs to R
then which of these holds true??
(a) f is many one
(b) f has no minima
(c) f is neither even nor odd
(d) f is bounded

please help me with this problem

Question

f(x) is a continuous fxn.
f(x)+f(2x+y)+5xy=f(3x-y)+2x^{2}+1 for all x belongs to R
then which of these holds true??
(a) f is many one
(b) f has no minima
(c) f is neither even nor odd
(d) f is bounded

please help me with this problem

anonymous · Answer

$\mathsf{f(x) = 1- \frac 1 2 x^2}$ satisfies the equation. But it's a even function, a parabola and has no defined Minima. I don't know what you mean by bounded, but I think it's probably not bounded. And because it's a parabola yes it's a many one function as well.
A and B.

anonymous · Answer

yeah thanxx i got it....thanx a lot..;)
u r really genious.:)

anonymous · Answer

You're welcome :-)

anonymous · Answer

but how this function came into ur mind...can u please tell??

anonymous · Answer

Umm first I defined y = 0. 
$$f(x) + (2x) - f(3x) = 2x^2 + 1\tag1$$
I checked the function for x=0
$$f(0) = 1$$
Then I just guessed a function, I really don't have much working to show. If f(0) =1, I presumed f(x) must be something like $ax^1 + bx^2 +\ldots +nx^n  + 1$. I used the equation (1) to get the idea the further idea of the function. I only needed to get one function that satisfies the equation $f(x)+f(2x+y)+5xy=f(3x-y)+2x^{2}+1$, for that I simply assumed the function to be $ax^2 +1$.