B. Let the function be
$$
y=(\sqrt{4+x}+\sqrt{4-x}+1)\left(\sqrt{16-x^{2}}+2\right) \text {. }
$$

Then the range of $y$ is . .
(A) $2 \leqslant y \leqslant 20$
(B) $2 \leqslant y \leqslant 30$
(C) $4 \sqrt{2}+2 \leqslant y \leqslant 20$
(D) $4 \sqrt{2}+2 \leqslant y \leqslant 30$
B. D.

Notice,
$$
\begin{array}{l}
y=\left[\sqrt{(\sqrt{4+x}+\sqrt{4-x})^{2}}+1\right]\left(\sqrt{16-x^{2}}+2\right) \\
=\left(\sqrt{8+2 \sqrt{16-x^{2}}}+1\right)\left(\sqrt{16-x^{2}}+2\right)
\end{array}
$$

Therefore, when $x=4$, $y_{\text {min }}=\sqrt{2}+2$;
when $x=0$, $y_{\max }=30$.
Hence, the range of $y$ is $4 \sqrt{2}+2 \leqslant y \leqslant 30$.