2018 AMC 12A 第 21 题

By Descartes’ Rule of Signs each polynomial has exactly one real root, and it lies in $(-1, 0)$

On $(-1, 0)$ a polynomial with larger values has a smaller root; note $x^{19} \gt x^{17}$ and $x^{13} \gt x^{11}$ there

Each polynomial in choices A–D has no positive root and exactly one negative root, which lies in $(-1, 0)$ (it is positive at $0$ and negative at $-1$) and is increasing there. On the interval $(-1, 0),$ $x^{19} \gt x^{17}$ and $x^{13} \gt x^{11}.$ Thus each of A, C, and D has a larger value than B at every point of this interval, so each crosses $0$ to the left of B. Therefore B has the greatest root among A–D.

The linear choice E has root $e=-\tfrac{2018}{2019}.$ Since $\tfrac{2018}{2019}\gt\tfrac9{10},$ we have $e^{17}+2018e^{11}+1$ $\lt-2018\left(\tfrac9{10}\right)^{11}+1$ $\lt0.$ Because B is increasing and is negative at $e,$ its root lies to the right of E’s. Hence B has the greatest real root.

Thus, the correct answer is B.