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Let $f$ be a polynomial satisfying $f(x^2)-xf(x) = x^4(x^2-1), x inBbb R^+$. Then which of the following is correct?

A) $f$ is an even function

B) $f$ is an odd function

C) $displaystylelim_{xto infty} frac{f(x)}{x^3}=1$

D) $displaystylelim_{xto infty} left(frac{f(x)}{x^2}-x right)$ exist and is equal to a non-zero quantity.

I have no idea what to do here.

Looking at the options, one thing I could guess is that the question wants us to find $f(x)$. After an analysis of few minutes, I could guess $f(x) = x^3$. But that gives me the answer as B), C). But the answer given is just C).

Any help would be appreciated.

Some questions have been raised in the comments about my answer below. A polynomial on $mathbb R$ which satisfies the given functional equation on $[0,infty)$ necessarily has the form $ax+x^{3}$ on that interval. But then it must have the same form on all of $mathbb R$. So I think the answer below is correct.

Look at the degree of the polynomial. If $f$ has degree $n$ then LHS has degree $2n$ and RHS has degree $6$. Hence $f$ is a polynomial of degree 3. The constant term is $0$ because $f(0)=0$. Show that the term in $x^{2}$ also must vanish because $xf(x)=f(x^{2})-x^{6}+x^{4}$ is a polynomial with only even powers of $x$. I guess you can take it from here. [All solutions are of the form $ax+x^{3}$]. Final anwser: A) and D) are false (the limit in D) exists but it is $0$); B) and C) are true.

You don't need to find $f(x)$. For instance, statements A and B can be dealt with by considering
$$
f(x)=frac{f(x^2)-x^4(x^2-1)}{x} tag{*}
$$
and so
$$
f(-x)=frac{f(x^2)-x^4(x^2-1)}{-x}=-f(x)
$$
showing that $f$ is an odd function. One can object that (*) only holds for $x>0$, but the right-hand side is a polynomial, (implying $f(0)=0$) and if two polynomials agree on an infinite set they're equal.

If the degree of $f$ is $n$, then the degree of $f(x^2)$ is $2n$. By (*), the degree of $f(x^2)$ must be $6$, or the equality could not hold. Hence $n=3$.

The function $f(x)/x^3$ has finite limit $l$, owing to $deg f(x)=3$. Now
$$
frac{f(x^2)-xf(x)}{x^6}=frac{f(x^2)}{(x^2)^3}-frac{1}{x^2}frac{f(x)}{x^3}=frac{x^2-1}{x^2}=1-frac{1}{x^2}
$$
Hence $l=1$.

Similarly,
$$
frac{f(x)}{x^2}-x=frac{f(x^2)}{x^3}-x(x^2-1)-x=frac{f(x^2)}{x^3}-x^3=
xleft(frac{f(x^2)}{(x^2)^2}-x^2right)
$$
Now it's clear that statement D is false: the limit exists (finite or infinite), but if it's finite it must be $0$.

Therefore B and C are true. This allows us to find $f(x)$: it is a degree $3$ polynomial, with leading coefficient $1$ and no term of even degree. Hence we have
$$
f(x)=x^3+ax
$$
Apply the functional equation:
$$
f(x^2)-xf(x)=x^6+ax^2-x^4-ax^2=x^4(x^2-1)
$$
holds for every $x$. Therefore $a$ can be anything.

To expand on @KaviRamaMurthy's answer:
Let $f=sum^n a_kx^k$ where $a_nne 0$ for $xge 0$. Then
$$sum a_kx^{2k}-sum a_kx^{k+1}=x^6-x^4$$
$$sum a_kx^{2k}=sum a_kx^{k+1}+x^6-x^4$$
So $2nin [0,n+1]cup{6}$ which means $nle 1$ or $n=3$. So
$$f=b_0+b_1x+b_2x^2+b_3x^3$$ for $xge 0$. And then
$$b_0+b_1x^2+b_2x^4+b_3x^6=b_0x+b_1x^2+b_2x^3+b_3x^4+x^6-x^4$$
which means $b_3=1$ and then $b_2=0$: $b_1$ is unconstrained and $b_0=0$:
$$f(x)=x^3+b_1x.$$

The only way (B) can hold is if they meant

for $xin R^+$, $f$ is equal to a polynomial such that... but $f$ need not be a polynomial on all of $R$.