$2 \cdot 113$ equation $x^{6}-3 x^{5}-6 x^{3}-x+8=0$
(A) No real roots.
(B) Exactly two distinct negative roots.
(C) Exactly one negative root.
(D) No negative roots, but at least one positive root.
(E) None of these.
(26th American High School Mathematics Examination, 1975)
[Solution] If $r$ is a root of the given equation, then $r^{6}+8=3 r^{5}+6 r^{3}+r$. If $r<0$, and $f(1)=-1<0$, it is known that $f(x)$ has at least one real root between $(0,1)$. Therefore, the answer is $(D)$.