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If two real polynomials $f(x)$ and $g(x)$ of degrees $m(\geq 2)$ and $n(\geq 1)$ respectively satisfy $$f(x^2+1)=f(x)g(x)$$ for every $x \in R$ then

(A) $f$ has exactly one real root $x_0$ such that $f'(x_0)\neq 0$

(B) $f$ has exactly one real root $x_0$ such that $f'(x_0)= 0$

(C) $f$ has $m$ distinct real roots

(D) $f$ has no real root

I tried finding value to this function and what I've figured out is that if $f(x)$ is of degree-2, then $g(x)$ should also be of degree-2 but I'm unable to find an equation of $f(x)$ where it solves $\frac{f(x^{2}+1)}{f(x)}=g(x)$

Thanks in advance!

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1 Answer 1

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Note that if $x_0$ is a root of $f(x)$, then so is $x_0^2+1$. Also note that for all reals $x$, it holds that $x^2+1>x$. Thus, if $x_0$ is a real root of $f(x)$, then $x_{n} = x_{n-1}^2+1$, $n\ge 1$ gives infinitely many real roots of $f(x)$, which implies that $f(x) = 0$. This leads to a contradiction, and we conclude there is no real root of $f(x)$.

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