Let $f(x)$ be a $n$ degree polynomial function having $n$ real and distinct roots. If $g(x) = f'(x) + 100f(x)$, then minimum number of roots that $g(x)$ must possess is:

$\text {a) n}$

$\text {b) n+1}$

$\text {c) n-1}$

$ \text {d) None of these}$

I don't really know what to do here. I assumed that since $g(x)$ is of degree $n$ it must have minimum $n$ roots. I cannot really describe the relationship among the equations.

The proposed solution (which I do not understand is): enter image description here

  • $\begingroup$ Note: it's hard to read, but the second line of the proposed solution should read $C'_1(x)=e^{100x}g(x)$. $\endgroup$
    – lulu
    Oct 20, 2018 at 16:42
  • 1
    $\begingroup$ Second note; I don't see how we can have exactly $n-1$ real roots. Since the coefficients are (obviously) real, the non-real solutions must occur in conjugate pairs...thus, if you have $n-1$ real roots you must have $n$ real roots. $\endgroup$
    – lulu
    Oct 20, 2018 at 16:44
  • $\begingroup$ @lulu The solution does read what you said. And the third line is $C'(x) = 0$ implies $g(x) = 0$. The solution doesn't say it has exactly $n-1$ real roots, it says that it has atleast $n-1$ real roots. $\endgroup$
    – Iceberry
    Oct 20, 2018 at 17:16
  • $\begingroup$ I understand. My point was that "at least $n-1$ real roots" implies "$n$ real roots" since you can not have exactly $n-1$. ( and on the type setting issue my point was that the derivative mark is illegible in the reproduction). $\endgroup$
    – lulu
    Oct 20, 2018 at 17:22
  • 2
    $\begingroup$ I agree with @lulu ; here's another reason. Certainly $g(x)$ has real coefficients and is degree $n$. But the constant coefficient of $g$ (up to a sign) is the product of its roots by Vieta's formulas, so if $n-1$ roots are real then the remaining root is real as well. (Although this is absolutely unneccessary given the complex pairs argument) EDIT: See Martin's comment below; you should use the sum of coefficients, not the product $\endgroup$
    – TomGrubb
    Oct 20, 2018 at 17:32

1 Answer 1


$C(x) = e^{100x} f(x)$ has $n$ distinct zeros $\alpha_1 < \alpha_2 < \ldots < \alpha_n$, therefore $$ C'(x) = e^{100x} \bigl(f'(x) + 100 f(x) \bigr) = e^{100x}g(x) $$ has (at least) one zero $x_k$ in each of the $n-1$ intervals $(\alpha_k, \alpha_{k+1})$,$1 \le k \le n-1$.

Also $\lim_{x \to -\infty} C(x) = 0$, therefore $C(x)$ has a local extremum at some point $x_0 \in (-\infty, \alpha_1)$, and $C'(x_0) = 0$.

This gives $n$ distinct real roots $x_0< x_1 < \ldots < x_{n-1}$of the polynomial $g$, and there cannot be more because of its degree.

  • $\begingroup$ The same argument shows that, for any $k\in\mathbb{R}$ and $f(x)\in\mathbb{R}[x]$, the number of real roots of $f(x)+k\,f'(x)$ is greater than or equal to the number of real roots of $f(x)$. Here, multiplicities are counted. (It is possible for $f(x)+k\,f'(x)$ to have more real roots than $f(x)$, though.) $\endgroup$ Oct 20, 2018 at 22:40

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