Test B Given the function $f(x)=a x^{2}-c$, satisfying $-4 \leqslant f(1) \leqslant-1,-1$ $\leqslant f(2) \leqslant 5$, then $f(3)$ should satisfy:
(A) $-7 \leqslant f(3) \leqslant 26$;
(B) $-4 \leqslant f(3) \leqslant 15$;
(C) $-1 \leqslant f(3) \leqslant 20$;
(D) $-\frac{28}{3} \leqslant f(3) \leqslant \frac{35}{3}$.
Noting that $f(x)$ is an even function, we have $f(-1)=f(1)$. By the Lagrange polynomial, we get
$$
\begin{aligned}
f(x)= & f(-1) \frac{(x-1)(x-2)}{(-1-1)(-1-2)}+f(1) \frac{(x+1)(x-2)}{(1+1)(1-2)} \\
& +f(2) \frac{(x+1)(x-1)}{(2+1)(2-1)} \\
= & -\frac{1}{3}\left(x^{2}-4\right) f(1)+\frac{1}{3}\left(x^{2}-1\right) f(2),
\end{aligned}
$$

Thus,
$$
f(3)=-\frac{5}{3} f(1)+\frac{8}{3} f(2) .
$$

Therefore, from the condition we get $-1 \leqslant f(3) \leqslant 20$. The answer is (C).