7・90 Let $f(x)=a x^{7}+b x^{3}+c x-5$, where $a, b$ and $c$ are constants. If $f(-7)=7$, then $f(7)$ equals
(A) -17.
(B) -7.
(C) 14.
(D) 21.
(E) cannot be uniquely determined.
(33rd American High School Mathematics Examination, 1982)
[Solution] Note the characteristics of the given function $f(x)$: the powers of $x$ are all odd, hence
$$
f(-x)=-a x^{7}-b x^{3}-c x-5,
$$

Therefore, $f(x)+f(-x)=-10$,
and $f(7)+f(-7)=-10$.
Given that $f(-7)=7$, it follows that $f(7)=-17$.
Thus, the answer is $(A)$.