Quartic polynomials having no real roots 
Let $$ f(n,x) = 1 + \dfrac{x}{2^n} + \dfrac{x^2}{3^n} + \dfrac{x^3}{4^n} + \dfrac{x^4}{5^n}$$ The value(s) of the positive integer $n$ such that $f(n, x) = 0$ has no real roots is/are

*

*A) $2021$

*B) $2022$

*C) $4$

*D) $7$


My attempt
I tried to compute the derivatives of $f(n, x)$ with respect to $x$ and found that the second derivative was strictly increasing for such values of $n$ given in the options. That gives me $f'(n,x)$ has only one real root. I am unable to observe anything about $f(n, x)$. Any constructive hint is appreciated.
 A: We could find that $ \frac{d xf(n,x)}{dx}=f(n-1,x)$
Since f(0,x)=1+x+x^2+x^3+x^4 has no real roots (4 primitive roots of $x^5=1$)
so $f(0,x)\gt 0$ for all real number x so that xf(1,x) is monotone increasing function and it has only a single real root x=0 so that f(1,x) has no real root so that $f(1,x) \gt 0$ for all real number x.
Similarly, by assuming $f(n-1,x)\gt 0$ for all real number, we could find that $xf(n,x)$ is monotone increasing function so that it has only a unique real root x=0.
so that $f(n,x) \gt 0$ for all real number 0.
A: Obviously, any roots are negative. Write $x = - y^n$, where $y > 0$. Then our equation takes the form
$$1 - (y/2)^n + (y^2/3)^n - (y^3/4)^n + (y^4/5)^n = 0.$$
If $y > 3/2$, then (in absolute value) the fifth term exceeds the fourth and the third exceeds the second.
If $5/4 < y < 2$ , then (in absolute value) the fifth term exceeds the fourth and the first exceeds the second.
If $y < 4/3$, then (in absolute value) the third term exceeds the fourth and the first exceeds the second.
Since any positive value of $y$ lies in one of these three intervals, the equation has no solutions.
A: It is clear that the quartic polynomial $ \ 1 + \dfrac{x}{2^n} + \dfrac{x^2}{3^n} + \dfrac{x^3}{4^n} + \dfrac{x^4}{5^n} \ \ $ has no positive real zeroes by the "Rule of Signs" (or even just "by inspection") and that any possible negative real zeroes would be due to the terms with odd powers of $ \ x \ $ .
However, we can look at the behavior of pairs of terms to show that the function $ \ f(n,x) \ $ is positive for all real $ \ x \ \ . $  For the second and third terms in ascending order,
$$ \frac{x^2}{3^n} \ + \ \frac{x}{2^n} \ \ = \ \ \frac{1}{3^n} · \left( \ x^2 \ + \ \frac{3^n}{2^n}·x \ \right) \ \ = \ \ \frac{1}{3^n} · \left( \ x \ + \ \frac12·\frac{3^n}{2^n} \ \right)^2 \ - \ \frac{3^n}{2^{2n+2}} \ \ . $$  The minimal value for this pair of terms as a function of $ \ n \ $ forms a geometric sequence with term ratio $ \ r \ = \ \frac{3}{2^2} \ \ , \ $ so the minimum has an absolute-value less than $ \ \dfrac{3^4}{2^{10}} \ = \ \dfrac{81}{1024} \ \ \approx \ \ 0.0791 \ \ $ for all natural numbers $ \ n \ \ge \ 4 \ \ . $
For the fourth and fifth terms in ascending order, we wish to establish the inequality
$$ \frac{x^4}{5^n} \ + \ \frac{x^3}{4^n} \ \ \ge \ \ \frac{x^4}{5^n} \ - \ \frac{x^2}{3^n} \ \ ; $$  the function on the right side of the inequality is an even-symmetry function and we will consider its behavior for $ \ x \ < \ 0 \ \ . $  From this, we have
$$ \frac{x^3}{4^n} \ + \ \frac{x^2}{3^n} \ \ = \ \ x^2 · \left(\frac{x }{4^n} \ + \ \frac{1}{3^n} \right) \ \ \ge \ \ 0 \ \  , $$
which holds in the interval $ \ -\left(\frac43 \right)^n \ \le \ x \ \le \ 0 \ \ . $ For the even-symmetry "comparison function", we obtain
$$ \frac{x^4}{5^n} \ - \ \frac{x^2}{3^n} \ \ = \ \ \frac{1}{5^n} · \left( \ x^4 \ - \ \frac{5^n}{3^n}·x^2 \ \right) \ \ = \ \ \frac{1}{5^n} · \left( \ x^2 \ - \ \frac12·\frac{5^n}{3^n} \ \right)^2 \ - \ \frac{5^n}{4·3^{2n}} \ \ . $$
(The squared-$x \ $ in the parenthetical factor is the result of the symmetry of the function.)  Since $ \ x \ = \ -\frac{1}{\sqrt2}·\left(\sqrt{\frac53} \right)^n \ $ is within the interval $ \ -\left(\frac43 \right)^n \ \le \ x \ \le \ 0 \ \  , $ we are able to say that
$$ \frac{x^4}{5^n} \ + \ \frac{x^3}{4^n} \ \ \ge \ \ - \ \frac{5^n}{4·3^{2n}} \ \ . $$
This minimum as a function of $ \ n \ $ forms a geometric sequence with term ratio $ \ r \ = \ \frac{5}{3^2} \ \ , \ $ so it has an absolute-value less than $ \ \dfrac{5^4}{4·3^8} \ = \ \dfrac{625}{26,244} \ \ \approx \ \ 0.0238 \ \ $ for all natural numbers $ \ n \ \ge \ 4 \ \ . $
Hence, $ \ f(n,x) \ = \   1 + \dfrac{x}{2^n} + \dfrac{x^2}{3^n} + \dfrac{x^3}{4^n} + \dfrac{x^4}{5^n} \ \ge \ 1 - 0.103 \ \ $ for all real $ \ x \ $ at all natural numbers $ \ n \ \ge \ 4 \ \ ; $ its four zeroes thus occur in two complex-conjugate pairs.
