2.43 Two prime numbers $a, b$ are exactly the roots of the integer coefficient equation $x^{2}-21 x+t=0$ for $x$. Then $\frac{b}{a}+\frac{a}{b}$ equals
(A) 2213.
(B) $\frac{58}{21}$.
(C) $\frac{2402}{49}$.
(D) $\frac{365}{38}$.
(2nd "Hope Cup" National Mathematics Invitational Competition, 1991)
[Solution] From the given and Vieta's formulas, we know: $a+b=21$, so $a=2, b=19$; or $a=19, b=2$.
Then $\frac{b}{a}+\frac{a}{b}=\frac{a^{2}+b^{2}}{a b}=\frac{2^{2}+19^{2}}{2 \cdot 19}=\frac{365}{38}$. Therefore, the answer is $(D)$.