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):
$C(x) = e^{100x} f(x)$ 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.