smc

Denote the polynomials in the answer choices by $A(x),B(x),C(x),D(x),$ and $E(x),$ respectively. Note that $A(x),B(x),C(x),D(x),$ and $E(x)$ are strictly increasing functions with range $(-\infty ,\infty ).$ So, each polynomial has exactly one real root. The real root of $E(x)$ is $x=-\frac{2018}{2019}\approx -\mathrm{1.000.}$ On the other hand, since $A(-1)=B(-1)=C(-1)=D(-1)=-2018$ and $A(0)=B(0)=C(0)=D(0)=1,$ we conclude that the real root for each of $A(x),B(x),C(x),$ and $D(x)$ must satisfy $x\in (-1,0)$ by the Intermediate Value Theorem (IVT). We analyze the polynomials for $x\in (-1,0):$ 1. We have $$\begin{array}{rl}B(x)-A(x)=D(x)-C(x)& =x^{17}-x^{19}\\ & =x^{17}\left(1-x^2\right)\\ & <0.\end{array}$$ As the graph of $y=A(x)$ is always above the graph of $y=B(x)$ in this interval, we deduce that $B(x)$ has a greater real root than $A(x)$ does. By the same reasoning, $D(x)$ has a greater real root than $C(x)$ does. 2. We have $$\begin{array}{rl}B(x)-D(x)& =2018x^{11}-2018x^{13}\\ & =2018x^{11}\left(1-x^2\right)\\ & <0,\end{array}$$ from which $B(x)$ has a greater real root than $D(x)$ does. Now, we are left with comparing the real roots of $B(x)$ and $E(x).$ Since $B\left(-\frac{1}{\sqrt{2}}\right)<0<B\left(-\frac{1}{2}\right),$ it follows that the real root of $B(x)$ must satisfy $x\in \left(-\frac{1}{\sqrt{2}},-\frac{1}{2}\right)$ by the IVT. Clearly, $\boxed{x^{17}+2018x^{11}+1}$ has the greatest real root.